explain plan for the same query diff

Similar Questions

• Why I have two different execution plans for the same query on two different servers

  Hello everyone.

  I need your help to solve the problem quickly.

  In a nutshell, we have two servers that have the same version of Oracle RDBMS (11.2.0.4 EE). One of them for purposes of development and another is a production one.

  We have therefore two different execution plans for the same query executed on both servers. The only case of execution is OK and another is too slow.

  So I have to slow down the work of query using the same scheme to explain that young.

  Fence wire.

• The execution plan changes for the same query.

  Hi all
  
  This issue was raised before also, but still not able to find the real cause of this.
  
  Thread1:
  Re: Research of fragmentation of the table in Oracle 8.1.6.3.0
  
  Thread2:
  CBC latch and buffer busy await you on the same table.
  
  It comes, sometimes hammers server 100% CPU utilization with free latch and buffer busy wait events.
  
  We found a single query consumes high CPU usage that is run by different sessions.
  
  This query have two types of execution plans, where one is accurate and is not (its primary key hit index index no appropriate means present on the table)
  
  Because its primary key index hit repeatedly at various sessions, some sessions are powerful db file sequential read and a few sessions waiting buffer busy waits for event. Also during this time a few sessions waiting for latch free event.
  
  My doubt is how to sql even with different literal values execution plan changes and causes a prob.
  

  select count(*),event from v$session_wait group by event;
    COUNT(*) EVENT
  ---------- ----------------------------------------------------------------
         165 SQL*Net message from client
           1 SQL*Net message to client
           3 buffer busy waits
           2 db file parallel read
          18 db file sequential read
          10 latch free
           5 log file sync
           1 pmon timer
           6 rdbms ipc message
           1 smon timer
  
  SQL> select sid from v$session_wait where event='db file sequential read';
         SID
  ----------
          26
          58
          82
         107
         116
         223
         212
         203
         192
         173
         161
         157
         150
         147
         254
         238
         229
         112
         101
          81
          68
  
  SQL> select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=&SID;
  Enter value for sid: 161
  old   1: select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=&SID
  new   1: select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=161
  
  SPID             SID    SERIAL# PROGRAM
  --------- ---------- ---------- ------------------------------------------------
  4231             161      49569 oracle@tfrdb3 (TNS V1-V3)
  
  
  SQL> select sql_text
  from v$process a,
       v$session b,  2    3
       v$sql c
  where a.addr = b.paddr and
       b.sql_hash_value = c.hash_value and
      a.spid = &PID;  4    5    6    7
  Enter value for pid: 4231
  old   7:     a.spid = &PID
  new   7:     a.spid = 4231
  
  SQL_TEXT
  --------------------------------------------------------------------------------
  SELECT ERROR,TIME_STAMP,O_RESOURCE,QUEUE,NEW_QUEUE FROM LOG WHERE ID = '09292AMR
  10B41FE' AND TYPE IN (11, 28, 25, 18, 60, 13) AND (LOG_SEQ>'234225222' OR TYPE =
   18 AND LOG_SEQ='234225222') ORDER BY TIME_STAMP ASC
  
  SQL> set autotrace traceonly exp
  SQL> SELECT ERROR,TIME_STAMP,O_RESOURCE,QUEUE,NEW_QUEUE FROM amrwf1.LOG WHERE ID = '09292AMR10B41FE' AND TYPE IN (11, 28, 25, 18, 60, 13) AND (LOG_SEQ>'234225222' OR TYPE =18 AND LOG_SEQ='234225222') ORDER BY TIME_STAMP ASC;
  
  Execution Plan
  ----------------------------------------------------------
     0      SELECT STATEMENT Optimizer=CHOOSE (Cost=11 Card=2 Bytes=126)
     1    0   SORT (ORDER BY) (Cost=11 Card=2 Bytes=126)
     2    1     CONCATENATION
     3    2       TABLE ACCESS (BY INDEX ROWID) OF 'LOG' (Cost=4 Card=1
            Bytes=63)
  
     4    3         INDEX (UNIQUE SCAN) OF 'PK_LOG_LOG_SEQ' (UNIQUE) (Co
            st=3 Card=1)
  
     5    2       TABLE ACCESS (BY INDEX ROWID) OF 'LOG' (Cost=4 Card=1
            Bytes=63)
  
     6    5         INDEX (RANGE SCAN) OF 'PK_LOG_LOG_SEQ' (UNIQUE) (Cos
            t=3 Card=1)
  
  
  SQL> select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=&SID;
  Enter value for sid: 147
  old   1: select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=&SID
  new   1: select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=147
  
  SPID             SID    SERIAL# PROGRAM
  --------- ---------- ---------- ------------------------------------------------
  6255             147      38306 oracle@tfrdb3 (TNS V1-V3)
  
  SQL> select sql_text
  from v$process a,
       v$session b,
       v$sql c  2    3
  where a.addr = b.paddr and
       b.sql_hash_value = c.hash_value and
      a.spid = &PID;  4    5    6    7
  Enter value for pid: 6255
  old   7:     a.spid = &PID
  new   7:     a.spid = 6255
  
  SQL_TEXT
  --------------------------------------------------------------------------------
  SELECT ERROR,TIME_STAMP,O_RESOURCE,QUEUE,NEW_QUEUE FROM LOG WHERE ID = '09273AMR
  62B4894' AND TYPE IN (11, 28, 25, 18, 60, 13) AND (LOG_SEQ>'223324996' OR TYPE =
   18 AND LOG_SEQ='223324996') ORDER BY TIME_STAMP ASC
  
  
  SQL> set autotrace traceonly exp
  SQL> SELECT ERROR,TIME_STAMP,O_RESOURCE,QUEUE,NEW_QUEUE FROM amrwf1.LOG WHERE ID = '09273AMR62B4894' AND TYPE IN (11, 28, 25, 18, 60, 13) AND (LOG_SEQ>'223324996' OR TYPE =18 AND LOG_SEQ='223324996') ORDER BY TIME_STAMP ASC;
  
  Execution Plan
  ----------------------------------------------------------
     0      SELECT STATEMENT Optimizer=CHOOSE (Cost=1538 Card=736 Bytes=
            46368)
  
     1    0   SORT (ORDER BY) (Cost=1538 Card=736 Bytes=46368)
     2    1     TABLE ACCESS (BY INDEX ROWID) OF 'LOG' (Cost=1527 Card=7
            36 Bytes=46368)
  
     3    2       INDEX (RANGE SCAN) OF 'LOG_ID' (NON-UNIQUE) (Cost=32 C
            ard=736)
  
  
  
  SQL> select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=&SID;
  Enter value for sid: 82
  old   1: select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=&SID
  new   1: select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=82
  
  SPID             SID    SERIAL# PROGRAM
  --------- ---------- ---------- ------------------------------------------------
  6172              82      45378 oracle@tfrdb3 (TNS V1-V3)
  
  
  SQL> select sql_text
  from v$process a,
       v$session b,
       v$sql c
  where a.addr = b.paddr and
       b.sql_hash_value = c.hash_value and  2
    3      a.spid = &PID;  4    5    6    7
  Enter value for pid: 6172
  old   7:     a.spid = &PID
  new   7:     a.spid = 6172
  
  SQL_TEXT
  --------------------------------------------------------------------------------
  INSERT INTO LOG (ID,TIME_STAMP,TYPE,ERROR,INSTANCE,RULE_NUM,RULE_TYPE,PRIORITY,F
  LAGS,NAME,BATCH,O_RESOURCE,QUEUE,NEW_QUEUE,SERVER,FORM,WORKSET) VALUES (:V001,:V
  002,11,0,0,3,1,0,1,:V003,:V004,:V005,:V006,:V007,:V008,:V009,:V010)
  
  INSERT INTO LOG (ID,TIME_STAMP,TYPE,ERROR,INSTANCE,RULE_NUM,RULE_TYPE,PRIORITY,F
  LAGS,NAME,BATCH,O_RESOURCE,QUEUE,NEW_QUEUE,SERVER,FORM,WORKSET) VALUES (:V001,:V
  002,11,0,0,3,1,0,1,:V003,:V004,:V005,:V006,:V007,:V008,:V009,:V010)
  
  INSERT INTO LOG (ID,TIME_STAMP,TYPE,ERROR,INSTANCE,RULE_NUM,RULE_TYPE,PRIORITY,F
  LAGS,NAME,BATCH,O_RESOURCE,QUEUE,NEW_QUEUE,SERVER,FORM,WORKSET) VALUES (:V001,:V
  002,11,0,0,3,1,0,1,:V003,:V004,:V005,:V006,:V007,:V008,:V009,:V010)
  
  SQL_TEXT
  --------------------------------------------------------------------------------
  
  INSERT INTO LOG (ID,TIME_STAMP,TYPE,ERROR,INSTANCE,RULE_NUM,RULE_TYPE,PRIORITY,F
  LAGS,NAME,BATCH,O_RESOURCE,QUEUE,NEW_QUEUE,SERVER,FORM,WORKSET) VALUES (:V001,:V
  002,11,0,0,3,1,0,1,:V003,:V004,:V005,:V006,:V007,:V008,:V009,:V010)
  
  INSERT INTO LOG (ID,TIME_STAMP,TYPE,ERROR,INSTANCE,RULE_NUM,RULE_TYPE,PRIORITY,F
  LAGS,NAME,BATCH,O_RESOURCE,QUEUE,NEW_QUEUE,SERVER,FORM,WORKSET) VALUES (:V001,:V
  002,11,0,0,3,1,0,1,:V003,:V004,:V005,:V006,:V007,:V008,:V009,:V010)
  
  INSERT INTO LOG (ID,TIME_STAMP,TYPE,ERROR,INSTANCE,RULE_NUM,RULE_TYPE,PRIORITY,F
  LAGS,NAME,BATCH,O_RESOURCE,QUEUE,NEW_QUEUE,SERVER,FORM,WORKSET) VALUES (:V001,:V
  
  SQL_TEXT
  --------------------------------------------------------------------------------
  002,11,0,0,3,1,0,1,:V003,:V004,:V005,:V006,:V007,:V008,:V009,:V010)
  
  INSERT INTO LOG (ID,TIME_STAMP,TYPE,ERROR,INSTANCE,RULE_NUM,RULE_TYPE,PRIORITY,F
  LAGS,NAME,BATCH,O_RESOURCE,QUEUE,NEW_QUEUE,SERVER,FORM,WORKSET) VALUES (:V001,:V
  002,11,0,0,3,1,0,1,:V003,:V004,:V005,:V006,:V007,:V008,:V009,:V010)

  How to avoid this... why its different execution plan using (I mean bad index PK)
  
  Is it possible to avoid this?
  
  If any details please check out some of my previous post on this specific URL (above)
  
  -Yasser

  My doubt is how to sql even with different literal values execution plan changes and causes a prob.

  Different literal values cause analysis difficult.
  Hard analysis includes the re-evaluation of the best path.
  Literal value is included in the assessment of the selectivity for the scan interval (log_seq >...)

  See
  http://www.centrexcc.com/A%20Look%20under%20The%20Hood%20Of%20CBO%20-%20The%2010053%20Event.ppt.PDF
  http://www.centrexcc.com/fallacies%20Of%20The%20Cost%20Based%20Optimizer.PDF
  more the book of Jonathan Lewis which other threads, I believe that you already have.

  You must lower your CPU.
  Previous discussions, if the situation is still the same, it sounded like hard analysis particularly with this SELECTION against the NEWSPAPER plays an important role in that.

  How to avoid this... why its different execution plan using (I mean bad index PK)

  The points raised in the previous discussion remain valid.
  -Do you have access to this SQL to change?
  for example using bind variable or trick it if necessary due to problems caused by data as discussed in the previous thread.
  - Or you could it repoint the view to a view and a hint?
  -If a particular user makes this sql, could affect you cursor_sharing just for this user. If not, you should consider implementing pan-Canadian database.

  Oracle 8.1.6 still?

• Different plan for the same sql id

  Hello

  Our team of application reported recently that a query ran longer than usual.

  Search in tables AWR (dba_hist_active_sess_history & dba_hist_sqlstat), it was a change of plan_hash_value. The next day, without intervention from anyone (collect statistics did not run, no change in the tables involved) the sql used the former plan and completed quickly.

  The differences in the plan was the index that was used to access the table.

  What could be the reason for the optimizer to choose a bad plan and return once more the good thing. ?

  The code sql that has been run twice & three in one day and then it years off cursor cache.

  Regime shifts are normal.

  Flipfloppping plans are quite normal.

  I'd be willing to bet that a large percentage of your SQL shows variations in implementation plans and you never notice.

  And there are several reasons why you might get a different plan.

  1 bind variable peeking - it is normal SQL for the age from the cache and then get analyzed once again with different bind variable leading to different cardinality estimates and plans

  2. the statistics change - it is normal that the statistics of change over time causing different cardinality estimates and the various plans, especially if the histograms are involved and changing buckets

  3. data change - especially if you use dynamic sampling it is normal to get a different sample of data leading to different cardinality estimates and plans

  4 SYSDATE - it's normal for queries with SYSDATE in them to recognize that time is never on leave and which, in conjunction with statistics or data, can lead to different cardinality estimates and plans

  5. integrated optimization features - it's normal for features like ACS to kick in and notice that they got forecasts wrong in an execution plan and force the recalculation to a different plan with an adjusted cardinality.

  Using DBMS_XPLAN. DISPLAY_CURSOR or DISPLAY_AWR you can get different plans (provided that they are in memory or AWR). The notes section should indicate cardinality comments have been a factor. You can also get links peeked with format mask of "+ PEEKED_BINDS".

• Different output for the same query in SQL Server and Oracle

  I have two tables table1 and table2
  
  -table1 has two columns c1 int and varchar c2. There are no constraints added in it. It contains data as shown below
  
  C1 c2
  -------------------
  d 6
  5 d
  102 g
  6%
  f 103
  5.
  501 j
  1 g
  601 n
  2 m
  
  -table2 has only a single column c1 int. There is no added in constraints. It contains data as shown below
  
  C1
  ----
  6
  1
  4
  3
  2
  
  now when I run below, given the query in sql server and oracle it gives me different results
  
  Select *.
  FROM table1
  table2 on table2.c1 = table1.c1 inner join (SELECT ROW_NUMBER() (any ORDER by ASC c1) AS c1 from table2)
  
  output of SQL server
  ------------------------
  C1 c2 c1
  --------------------------------
  1 g 1
  2 m 2
  3 h 3
  4 g 4
  5 d 5
  
  
  release of Oracle
  ----------------------
  C1 C2 C1
  ---------------------------------
  5 d 5
  4 g 4
  3 h 3
  1 g 1
  2 m 2
  
  
  If you notice the first column in the two outputs. It is sorted in sql server and no oracle.
  
  Why he behaves differently in oracle? Is there a way I can fix this problem in oracle?
  
  Thank you
  Jigs

  It is NOT a behavior "differently" in Oracle; you did not specify just an order that you expect of your results, so you'll get output in what order the fantasies of the database showing (ie. no guaranteed order). It is an artifact of the way the database chooses to collect the data and databases (or same sets of data within the same database) can and will most likely behave differently.

  Same SQL Server will not be guaranteed to always get your data in an orderly manner if you exclude the order by clause, even if you think that there always display the data in an orderly manner.

  Your solution is to add an order by clause, in the TWO databases, to force the order of the data output.

• I would spend a "plan of photography" "plan of Applications." But I did have the 'photography plan' for a year yet. Can I still sign up for the same year or is this a new subscription?

  I would spend a "plan of photography" "plan of Applications." But I did have the 'photography plan' for a year yet. Can I still sign up for the same year or is this a new subscription?

  In addition to the other suggestions here, you can also consult the plan switch option (described here):
  Update or change your plan of Adobe Creative Cloud

• Different LOVs of af: query and af:form for the same attribute of VO

  Hello
  
  We need show different LOVs af:query and af:form for the same attribute in VO.
  
  Is it possible to use LOV Switcher for this?
  
  Can how we use in the LOV Switcher attribute to check if she is seen Critearia line or VO?

  Please see this post - http://jobinesh.blogspot.com/2011/04/identifying-request-for-lov-from-search.html

• Sales order form, line status code is displayed as "WITHDRAWN". but the same code appears as "AWAITING_SHIPPING" for the same line in the database. can you please explain, where the value in frontend comes? It differs from the flow_status_code in oe_or

  Sales order form, line status code is displayed as "WITHDRAWN". but the same code appears as "AWAITING_SHIPPING" for the same line in the database.

  Where the value in the frontend is filling?

  It differs from the flow_status_code in oe_order_lines_all?

  Thanks for the correction.

  He slipped right out of my mind!

  Here is a detailed explanation of this case:

  Status of orders picked and picked partial

  Kind regards

  Bashar

• Multiple values for the same column in the columns of diffétent in the same row?

  Hi all
  I wonder how you can display different values for the same column in different columns on the same line. For example, using a CASE statement, I can:
  
  CASE WHEN CODE IN ('1 ', ' 3') THEN COUNT (ID) AS 'Y '.
  CASE WHEN CODE NOT IN ('1 ', am') THEN COUNT (ID) AS "N".
  
  Yes, that will produce two columns but will produce null values to empty and also two separate registers.
  
  
  Any ideas?
  
  Thank you

  Are you sure that this code works for you?

  Select ID
           ,CASE WHEN MODE_CODE IN ('1', '3') THEN COUNT( No) END as "Fulltime"
           ,CASE WHEN MODE_CODE NOT IN ('1', '3') THEN COUNT( No ) END  as "Other"
  From table
  group by ID
  

  I guess the code above fails because MODE_CODE is not in your group by?

  My suggestion would be to put the CASE in the COUNT:

  Select ID
           ,COUNT(CASE WHEN MODE_CODE IN ('1', '3') THEN No END) as "Fulltime"
           ,COUNT(CASE WHEN MODE_CODE NOT IN ('1', '3') THEN No END)  as "Other"
  From table
  group by ID
  

  CASE expressions return no. when the respective conditions are true and NULL otherwise.
  COUNTY will have non-null values.

• How can I get a list of jobs that are planned for the days of the future

  How can I get a list of jobs that are planned for the days of the future?

  In a previous article, I found a query that lists the scheduled tasks that are already running. Can I get a similar request for the future days which is compiled?

  Hi sreedevir,

  SELECT jobmst.jobmst_prntname, jobmst.jobmst_name, jobrun.*

  FROM jobrun JOIN jobmst ON jobmst.jobmst_id = jobrun.jobmst_id -joining tables jobmst and jobrun

  WHERE jobrun_proddt > = dateadd (dd, 1, datediff (dd, getdate())) 0, -future dates

  AND jobmst.jobmst_type = 2 -given jobs (and not groups)

  ORDER BY 1, 2 -Sort by name of the parent, then by task name

  Feel free to make any changes to your reporting needs

  ARO

  The Derrick

• So, I have creative cloud... It came with Photoshop and Lightroom... could I uninstall Lightroom and install Illustrator for the same price?

  It came with Photoshop and Lightroom... could I uninstall Lightroom and install Illustrator for the same price?

  No, you can not install Illustrator instead of Lightroom. You can buy Adobe Illustrator creative Cloud app unique: pricing and membership creative cloud plans | Adobe Creative Cloud

• Difference in the number of records for the same date - 11 GR 2

  Guy - 11 GR on Windows2005 2, 64-bit.
  
  BILLING_RECORD_KPN_ESP - is a monthly partitioned table.
  BILLING_RECORD_IDX #DATE - is a local index on "charge_date" in the table above.
  
  SQL > select / * + index (BILLING_RECORD_KPN_ESP BILLING_RECORD_IDX #DATE) * /.
  2 (trunc (CHARGE_DATE)) CHARGE_DATE;
  3 count (1) Record_count
  4. IN "RATOR_CDR". "" BILLING_RECORD_KPN_ESP ".
  where the 5 CHARGE_DATE = January 20, 2013.
  Group 6 by trunc (CHARGE_DATE)
  5 m
  
  CHARGE_DATE RECORD_COUNT
  ------------------ ------------
  2401 20 January 13-> > some records here.
  
  -> > Here I can see only '2041' records for Jan/20. But in the query below, it shows "192610" for the same date.
  
  Why is this difference in the number of records?
  
  SQL > select / * + index (BILLING_RECORD_KPN_ESP BILLING_RECORD_IDX #DATE) * /.
  (trunc (CHARGE_DATE)) CHARGE_DATE,
  2 count (1) Record_count
  3. FOR "RATOR_CDR." "" BILLING_RECORD_KPN_ESP ".
  "4 where CHARGE_DATE > 20 January 2013."
  Group of 5 by trunc (CHARGE_DATE)
  6 order by trunc (CHARGE_DATE)
  5 m
  
  CHARGE_DATE RECORD_COUNT
  ------------------ ------------
  192610 20 January 13-> > more records here
  JANUARY 21, 13 463067
  JANUARY 22, 13 520041
  23 JANUARY 13 451212
  JANUARY 24, 13 463273
  JANUARY 25, 13 403276
  JANUARY 26, 13 112077
  27 JANUARY 13 10478
  28 JANUARY 13 39158
  
  Thank you!

  Because in the second example you also select rows that have a nonzero component.

  The first example selects only rows that are 00:00:00

  (by the way, you should ask questions like this in the forum SQL)

• Determine the maximum length of a column and its use in the same query?

  I would like to determine the maximum length of a column and use it in the same query. Is this possible? IE:
  
  SELECT RPAD (last_name, SELECT MAX (LENGTH (last_name)) + 5 FROM user, '-')
  OF the user.
  
  Thank you

  Hello

  Welcome to the forum!

  jfraley wrote:
  I would like to determine the maximum length of a column and use it in the same query. Is this possible? IE:

  SELECT RPAD (last_name, SELECT MAX (LENGTH (last_name)) + 5 FROM user, '-')
  OF the user.

  Thank you

  Sure. You almost go in your message: a scalar subquery . Just put brackets around the subquery:

  SELECT      RPAD ( last_name
            , (
               SELECT  MAX (LENGTH (last_name)) + 5
               FROM    user_tbl          -- USER is not a good table name
              )
            , '-'
            )          AS padded_last_name
  FROM      user_tbl
  ;
  

  Scalar subqueries in SQL are like tape in your garage in canvas: they are used for thousands of different things and open, maybe 1% of them. Usually, there are better ways to achieve the same results, such as the analytical functions:

  SELECT      RPAD ( last_name
            , 5 + MAX (LENGTH (last_name)) OVER ()
            , '-'
            )          AS padded_last_name
  FROM      user_tbl
  ;
  

  Still another way is to make the subquery and then join his game as if it were a table of results:

  WITH     got_max_length     AS
  (
       SELECT     MAX (LENGTH (last_name))     AS max_length
       FROM     user_tbl
  )
  SELECT      RPAD ( u.last_name
            , m.max_length + 5
            , '-'
            )          AS padded_last_name
  FROM           user_tbl     u
  CROSS JOIN     got_max_length     m
  ;
  

  Published by: Frank Kulash, December 12, 2011 21:17
  Added cross join example

• Several numbers of cell for the same person

  Is it possible to store multiple numbers of cell for the same person in my contacts? For example: I have a contact with two phone numbers, these are all cell phones, but in the contacts that they marked as the default cell and home numbers. When I am trying to change home to cell number and record this is not registered. It is always the cell and home numbers. So my question is: how to change all the numbers to cell Type?

  When you change the contact, press the label, scroll and choose custom label. You can't have two identical labels, but you can have labels custom as mobile2, mobile3, etc.

• How I find myself with two passwords to iCloud for the same account?

  How I find myself with two passwords to iCloud for the same account?

  For security reasons, I only use iCloud for Contacts and "find my iPhone".

  Several months ago Apple forced me to change my password to iCloud.  So I did this.   In the last months, I used successfully the new password to 10 or 12 times when asked without problem.

  However, I received a notification of a software update for my Apple Watch. I did the update without problem. However, after completing the update my iPhone asked me to connect to my iCloud account. When I did this, she rejected my password. I made 4 attempts typing very slowly and with care for you sure I did it right.

  Then, operating under a hunch, I decided to enter my old password to iCloud. It worked?  I'm confused about this, any ideas would be appreciated.

  I had a similar problem last year. I had to call Apple. Contact is a bottom of this page.