Hi,

I have a task at hand to reduce the time taken for search query to
execute. The query fetches records which will have to sorted by
degrees away from the logged in user. I have a function which
calculates the degrees, but using this in the search query slows the
execution and takes about 10 secs to complete which is unacceptable.

Please advice. Your help is much appreciated

For more details plz see:

http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=97021


Thanks
Isfaar

On Feb 26, 7:41*am, paankh... (AT) googlemail (DOT) com wrote:
Hi Isfaar,
Would it be possible for you to post your query here along with udf
definition. When a scalar UDF is invoked for each row in a query, it
can degrade the performance i.e. for each row you incur UDF invocation
cost. An inline expression might be faster than UDF in this case.

Depending on your SQL Server version, you might have to adopt
different approach. If you are on SQL Server 2005, then cross apply
operator in conjuction with inline table function will do the trick
for you. Inline table function behaves differently than scalar
function.

If you are on SQL Server 2000, a precisely written subquery might do
the trick.

HTH,

Najm

On Feb 26, 7:41*am, paankh... (AT) googlemail (DOT) com wrote:
Hi Isfaar,
Would it be possible for you to post your query here along with udf
definition. When a scalar UDF is invoked for each row in a query, it
can degrade the performance i.e. for each row you incur UDF invocation
cost. An inline expression might be faster than UDF in this case.

Depending on your SQL Server version, you might have to adopt
different approach. If you are on SQL Server 2005, then cross apply
operator in conjuction with inline table function will do the trick
for you. Inline table function behaves differently than scalar
function.

If you are on SQL Server 2000, a precisely written subquery might do
the trick.

HTH,

Najm

On Feb 26, 7:41*am, paankh... (AT) googlemail (DOT) com wrote:
Hi Isfaar,
Would it be possible for you to post your query here along with udf
definition. When a scalar UDF is invoked for each row in a query, it
can degrade the performance i.e. for each row you incur UDF invocation
cost. An inline expression might be faster than UDF in this case.

Depending on your SQL Server version, you might have to adopt
different approach. If you are on SQL Server 2005, then cross apply
operator in conjuction with inline table function will do the trick
for you. Inline table function behaves differently than scalar
function.

If you are on SQL Server 2000, a precisely written subquery might do
the trick.

HTH,

Najm

I can only repeat the same as Hugo and Najm, it is hard to say anything
without seeing the actual table, indexes, sample data, required results, and
business requirements/rules. Reading all those posts in the link at
sqlteam.com I can see only sample code provided by Peso (Peter Larsson). The
problem could be difficult or easy to solve, but hard to say without seeing
the real requirements.

Perhaps some things to consider are using materialized path or nested sets
to stored the hierarchy of contacts. Those models provide very efficient
retrieval of distance between nodes info, but have more difficult methods
for maintaining data.

Here is just a small sample (using the sample table and data provided by
Peso) on how materialized path may look:

-- Sample table with data
CREATE TABLE Contacts (
c_from CHAR(1),
c_to CHAR(1),
PRIMARY KEY (c_from, c_to));

INSERT INTO Contacts
SELECT 'A', 'B' UNION ALL
SELECT 'B', 'D' UNION ALL
SELECT 'C', 'A' UNION ALL
SELECT 'C', 'E' UNION ALL
SELECT 'G', 'C' UNION ALL
SELECT 'B', 'G' UNION ALL
SELECT 'F', 'D' UNION ALL
SELECT 'E', 'F';

-- Table to store paths
CREATE TABLE Paths (c_path VARCHAR(200) PRIMARY KEY);

This is the real hurdle, recalculating all paths

-- Recursive CTE to populate the paths
WITH PathCTE
AS
(SELECT c_from, c_to,
CAST('.' + CAST(c_from AS VARCHAR(8)) + '.' +
CAST(c_to AS VARCHAR(8)) + '.' AS VARCHAR(200)) AS c_path
FROM Contacts AS C1
UNION ALL
SELECT C.c_from, C.c_to,
CAST(P.c_path + C.c_to + '.' AS VARCHAR(200))
FROM PathCTE AS P
JOIN Contacts AS C
ON P.c_to = C.c_from
WHERE P.c_path NOT LIKE '%.' +
CAST(C.c_from AS VARCHAR(10)) +
'.' +
CAST(C.c_to AS VARCHAR(10)) +
'.%')
INSERT INTO Paths
SELECT c_path FROM PathCTE;

-- Show all paths between B and D
SELECT c_path
FROM Paths
WHERE c_path LIKE '.B.%'
AND c_path LIKE '%.D.';

-- Shortest path distance, longest path distance, and number of paths
SELECT MIN(LEN(c_path) - LEN(REPLACE(c_path, '.', '')) - 2) AS
shortest_distance,
MAX(LEN(c_path) - LEN(REPLACE(c_path, '.', '')) - 2) AS
longest_distance,
COUNT(*) AS paths_cnt
FROM Paths
WHERE c_path LIKE '.B.%'
AND c_path LIKE '%.D.';

Looking at the paths found:
..B.D.
..B.G.C.A.B.D.
..B.G.C.E.F.D.

You may notice the second path reused the first path to reach the
destination. But this could be desired or not, again hard to say with no
requirements. Easy to handle but did not bother...

As you can see calculating the distance is easy, but maintenance offsets
that. Based on your needs and model, if data is static this may do.

HTH,

Plamen Ratchev
http://www.SQLStudio.com

I can only repeat the same as Hugo and Najm, it is hard to say anything
without seeing the actual table, indexes, sample data, required results, and
business requirements/rules. Reading all those posts in the link at
sqlteam.com I can see only sample code provided by Peso (Peter Larsson). The
problem could be difficult or easy to solve, but hard to say without seeing
the real requirements.

Perhaps some things to consider are using materialized path or nested sets
to stored the hierarchy of contacts. Those models provide very efficient
retrieval of distance between nodes info, but have more difficult methods
for maintaining data.

Here is just a small sample (using the sample table and data provided by
Peso) on how materialized path may look:

-- Sample table with data
CREATE TABLE Contacts (
c_from CHAR(1),
c_to CHAR(1),
PRIMARY KEY (c_from, c_to));

INSERT INTO Contacts
SELECT 'A', 'B' UNION ALL
SELECT 'B', 'D' UNION ALL
SELECT 'C', 'A' UNION ALL
SELECT 'C', 'E' UNION ALL
SELECT 'G', 'C' UNION ALL
SELECT 'B', 'G' UNION ALL
SELECT 'F', 'D' UNION ALL
SELECT 'E', 'F';

-- Table to store paths
CREATE TABLE Paths (c_path VARCHAR(200) PRIMARY KEY);

This is the real hurdle, recalculating all paths

-- Recursive CTE to populate the paths
WITH PathCTE
AS
(SELECT c_from, c_to,
CAST('.' + CAST(c_from AS VARCHAR(8)) + '.' +
CAST(c_to AS VARCHAR(8)) + '.' AS VARCHAR(200)) AS c_path
FROM Contacts AS C1
UNION ALL
SELECT C.c_from, C.c_to,
CAST(P.c_path + C.c_to + '.' AS VARCHAR(200))
FROM PathCTE AS P
JOIN Contacts AS C
ON P.c_to = C.c_from
WHERE P.c_path NOT LIKE '%.' +
CAST(C.c_from AS VARCHAR(10)) +
'.' +
CAST(C.c_to AS VARCHAR(10)) +
'.%')
INSERT INTO Paths
SELECT c_path FROM PathCTE;

-- Show all paths between B and D
SELECT c_path
FROM Paths
WHERE c_path LIKE '.B.%'
AND c_path LIKE '%.D.';

-- Shortest path distance, longest path distance, and number of paths
SELECT MIN(LEN(c_path) - LEN(REPLACE(c_path, '.', '')) - 2) AS
shortest_distance,
MAX(LEN(c_path) - LEN(REPLACE(c_path, '.', '')) - 2) AS
longest_distance,
COUNT(*) AS paths_cnt
FROM Paths
WHERE c_path LIKE '.B.%'
AND c_path LIKE '%.D.';

Looking at the paths found:
..B.D.
..B.G.C.A.B.D.
..B.G.C.E.F.D.

You may notice the second path reused the first path to reach the
destination. But this could be desired or not, again hard to say with no
requirements. Easy to handle but did not bother...

As you can see calculating the distance is easy, but maintenance offsets
that. Based on your needs and model, if data is static this may do.

HTH,

Plamen Ratchev
http://www.SQLStudio.com

I can only repeat the same as Hugo and Najm, it is hard to say anything
without seeing the actual table, indexes, sample data, required results, and
business requirements/rules. Reading all those posts in the link at
sqlteam.com I can see only sample code provided by Peso (Peter Larsson). The
problem could be difficult or easy to solve, but hard to say without seeing
the real requirements.

Perhaps some things to consider are using materialized path or nested sets
to stored the hierarchy of contacts. Those models provide very efficient
retrieval of distance between nodes info, but have more difficult methods
for maintaining data.

Here is just a small sample (using the sample table and data provided by
Peso) on how materialized path may look:

-- Sample table with data
CREATE TABLE Contacts (
c_from CHAR(1),
c_to CHAR(1),
PRIMARY KEY (c_from, c_to));

INSERT INTO Contacts
SELECT 'A', 'B' UNION ALL
SELECT 'B', 'D' UNION ALL
SELECT 'C', 'A' UNION ALL
SELECT 'C', 'E' UNION ALL
SELECT 'G', 'C' UNION ALL
SELECT 'B', 'G' UNION ALL
SELECT 'F', 'D' UNION ALL
SELECT 'E', 'F';

-- Table to store paths
CREATE TABLE Paths (c_path VARCHAR(200) PRIMARY KEY);

This is the real hurdle, recalculating all paths

-- Recursive CTE to populate the paths
WITH PathCTE
AS
(SELECT c_from, c_to,
CAST('.' + CAST(c_from AS VARCHAR(8)) + '.' +
CAST(c_to AS VARCHAR(8)) + '.' AS VARCHAR(200)) AS c_path
FROM Contacts AS C1
UNION ALL
SELECT C.c_from, C.c_to,
CAST(P.c_path + C.c_to + '.' AS VARCHAR(200))
FROM PathCTE AS P
JOIN Contacts AS C
ON P.c_to = C.c_from
WHERE P.c_path NOT LIKE '%.' +
CAST(C.c_from AS VARCHAR(10)) +
'.' +
CAST(C.c_to AS VARCHAR(10)) +
'.%')
INSERT INTO Paths
SELECT c_path FROM PathCTE;

-- Show all paths between B and D
SELECT c_path
FROM Paths
WHERE c_path LIKE '.B.%'
AND c_path LIKE '%.D.';

-- Shortest path distance, longest path distance, and number of paths
SELECT MIN(LEN(c_path) - LEN(REPLACE(c_path, '.', '')) - 2) AS
shortest_distance,
MAX(LEN(c_path) - LEN(REPLACE(c_path, '.', '')) - 2) AS
longest_distance,
COUNT(*) AS paths_cnt
FROM Paths
WHERE c_path LIKE '.B.%'
AND c_path LIKE '%.D.';

Looking at the paths found:
..B.D.
..B.G.C.A.B.D.
..B.G.C.E.F.D.

You may notice the second path reused the first path to reach the
destination. But this could be desired or not, again hard to say with no
requirements. Easy to handle but did not bother...

As you can see calculating the distance is easy, but maintenance offsets
that. Based on your needs and model, if data is static this may do.

HTH,

Plamen Ratchev
http://www.SQLStudio.com

On Feb 28, 8:51*am, "Plamen Ratchev" <Pla... (AT) SQLStudio (DOT) com> wrote:
Sorry for not having to replied earlier.

here is the code.

this is the query that gets fired;

SELECT Users.UserID, Users.FirstName + ' ' + Users.LastName as Name,
UserProfile.PropertyValue as Location,
professionalInfo.headline as Headline, industries.industryName as
Industry, professionalInfo.summary as Summary,
professionalInfo.interests, dbo.GetConnectionsCount(Users.UserID) AS
Connections,
dbo.GetRecommendations(Users.UserID) AS Recommendations,
dbo.fnCommonFriendsStep(Users.UserID, 36) AS Degree FROM Users
INNER JOIN UserProfile ON Users.UserID = UserProfile.UserID
INNER JOIN professionalInfo ON Users.UserID =
professionalInfo.memberId
INNER JOIN industries ON professionalInfo.primaryIndustry =
industries.industryId
WHERE (Users.UserID IN (SELECT DISTINCT UserID FROM vw_search ))
AND UserProfile.PropertyDefinitionID=29 AND
dbo.fnCommonFriendsStep(Users.UserID,36) >=0 ORDER By Degree ASC


fnCommonFriendstep one calculates relationship between the loggedin
member and other members.

if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].
[fnCommonFriendsStep]') and xtype in (N'FN', N'IF', N'TF'))
drop function [dbo].[fnCommonFriendsStep]
GO

SET QUOTED_IDENTIFIER ON
GO
SET ANSI_NULLS ON
GO

CREATE FUNCTION dbo.fnCommonFriendsStep
(
@Want1 INT,
@Want2 INT
)
RETURNS INT
AS

BEGIN
IF @Want1 = @Want2
RETURN 0

DECLARE @Friends TABLE (Generation INT, p INT)
DECLARE @Generation INT

SELECT @Generation = 0

INSERT @Friends
(
Generation,
p
)
SELECT 0,
memberId
FROM network
WHERE friendId = @Want1
UNION
SELECT 0,
friendId
FROM network
WHERE memberId = @Want1

WHILE NOT EXISTS (SELECT 1 FROM @Friends WHERE p = @Want2) AND
@Generation >= 0
BEGIN
SELECT @Generation = @Generation + 1

INSERT @Friends
(
Generation,
p
)
SELECT @Generation,
memberId
FROM network
WHERE friendId IN (SELECT p FROM @Friends WHERE Generation =
@Generation - 1)
AND memberId NOT IN (SELECT p FROM @Friends)
UNION
SELECT @Generation,
friendId
FROM network
WHERE memberId IN (SELECT p FROM @Friends WHERE Generation =
@Generation - 1)
AND friendId NOT IN (SELECT p FROM @Friends)

IF @@ROWCOUNT = 0
SELECT @Generation = -2
END

RETURN @Generation + 1
END
GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS ON
GO


GetRecommendations get the number of recommendations a member has
received.

if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].
[GetRecommendations]') and xtype in (N'FN', N'IF', N'TF'))
drop function [dbo].[GetRecommendations]
GO

SET QUOTED_IDENTIFIER ON
GO
SET ANSI_NULLS ON
GO

CREATE FUNCTION dbo.GetRecommendations
(
@user_id as int
)
RETURNS INT
AS
BEGIN
DECLARE @recomm INT
SELECT @recomm=Count(*) FROM endorsements WHERE memberId=@user_id AND
status='Accepted'
RETURN @recomm
END
GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS ON
GO


GetConnectionsCount fetches number of connections a member has


if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].
[GetConnectionsCount]') and xtype in (N'FN', N'IF', N'TF'))
drop function [dbo].[GetConnectionsCount]
GO

SET QUOTED_IDENTIFIER ON
GO
SET ANSI_NULLS ON
GO

CREATE FUNCTION dbo.GetConnectionsCount
(
@user_id as int
)
RETURNS INT
AS
BEGIN
DECLARE @recCount INT
SELECT @recCount=COUNT(*) FROM network WHERE (memberId=@user_id OR
friendId=@user_id) AND status=1
RETURN @recCount
END
GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS ON
GO


Please do let me know if you need any more information.

On Feb 28, 8:51*am, "Plamen Ratchev" <Pla... (AT) SQLStudio (DOT) com> wrote:
Sorry for not having to replied earlier.

here is the code.

this is the query that gets fired;

SELECT Users.UserID, Users.FirstName + ' ' + Users.LastName as Name,
UserProfile.PropertyValue as Location,
professionalInfo.headline as Headline, industries.industryName as
Industry, professionalInfo.summary as Summary,
professionalInfo.interests, dbo.GetConnectionsCount(Users.UserID) AS
Connections,
dbo.GetRecommendations(Users.UserID) AS Recommendations,
dbo.fnCommonFriendsStep(Users.UserID, 36) AS Degree FROM Users
INNER JOIN UserProfile ON Users.UserID = UserProfile.UserID
INNER JOIN professionalInfo ON Users.UserID =
professionalInfo.memberId
INNER JOIN industries ON professionalInfo.primaryIndustry =
industries.industryId
WHERE (Users.UserID IN (SELECT DISTINCT UserID FROM vw_search ))
AND UserProfile.PropertyDefinitionID=29 AND
dbo.fnCommonFriendsStep(Users.UserID,36) >=0 ORDER By Degree ASC


fnCommonFriendstep one calculates relationship between the loggedin
member and other members.

if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].
[fnCommonFriendsStep]') and xtype in (N'FN', N'IF', N'TF'))
drop function [dbo].[fnCommonFriendsStep]
GO

SET QUOTED_IDENTIFIER ON
GO
SET ANSI_NULLS ON
GO

CREATE FUNCTION dbo.fnCommonFriendsStep
(
@Want1 INT,
@Want2 INT
)
RETURNS INT
AS

BEGIN
IF @Want1 = @Want2
RETURN 0

DECLARE @Friends TABLE (Generation INT, p INT)
DECLARE @Generation INT

SELECT @Generation = 0

INSERT @Friends
(
Generation,
p
)
SELECT 0,
memberId
FROM network
WHERE friendId = @Want1
UNION
SELECT 0,
friendId
FROM network
WHERE memberId = @Want1

WHILE NOT EXISTS (SELECT 1 FROM @Friends WHERE p = @Want2) AND
@Generation >= 0
BEGIN
SELECT @Generation = @Generation + 1

INSERT @Friends
(
Generation,
p
)
SELECT @Generation,
memberId
FROM network
WHERE friendId IN (SELECT p FROM @Friends WHERE Generation =
@Generation - 1)
AND memberId NOT IN (SELECT p FROM @Friends)
UNION
SELECT @Generation,
friendId
FROM network
WHERE memberId IN (SELECT p FROM @Friends WHERE Generation =
@Generation - 1)
AND friendId NOT IN (SELECT p FROM @Friends)

IF @@ROWCOUNT = 0
SELECT @Generation = -2
END

RETURN @Generation + 1
END
GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS ON
GO


GetRecommendations get the number of recommendations a member has
received.

if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].
[GetRecommendations]') and xtype in (N'FN', N'IF', N'TF'))
drop function [dbo].[GetRecommendations]
GO

SET QUOTED_IDENTIFIER ON
GO
SET ANSI_NULLS ON
GO

CREATE FUNCTION dbo.GetRecommendations
(
@user_id as int
)
RETURNS INT
AS
BEGIN
DECLARE @recomm INT
SELECT @recomm=Count(*) FROM endorsements WHERE memberId=@user_id AND
status='Accepted'
RETURN @recomm
END
GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS ON
GO


GetConnectionsCount fetches number of connections a member has


if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].
[GetConnectionsCount]') and xtype in (N'FN', N'IF', N'TF'))
drop function [dbo].[GetConnectionsCount]
GO

SET QUOTED_IDENTIFIER ON
GO
SET ANSI_NULLS ON
GO

CREATE FUNCTION dbo.GetConnectionsCount
(
@user_id as int
)
RETURNS INT
AS
BEGIN
DECLARE @recCount INT
SELECT @recCount=COUNT(*) FROM network WHERE (memberId=@user_id OR
friendId=@user_id) AND status=1
RETURN @recCount
END
GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS ON
GO


Please do let me know if you need any more information.

On Feb 28, 8:51*am, "Plamen Ratchev" <Pla... (AT) SQLStudio (DOT) com> wrote:
Sorry for not having to replied earlier.

here is the code.

this is the query that gets fired;

SELECT Users.UserID, Users.FirstName + ' ' + Users.LastName as Name,
UserProfile.PropertyValue as Location,
professionalInfo.headline as Headline, industries.industryName as
Industry, professionalInfo.summary as Summary,
professionalInfo.interests, dbo.GetConnectionsCount(Users.UserID) AS
Connections,
dbo.GetRecommendations(Users.UserID) AS Recommendations,
dbo.fnCommonFriendsStep(Users.UserID, 36) AS Degree FROM Users
INNER JOIN UserProfile ON Users.UserID = UserProfile.UserID
INNER JOIN professionalInfo ON Users.UserID =
professionalInfo.memberId
INNER JOIN industries ON professionalInfo.primaryIndustry =
industries.industryId
WHERE (Users.UserID IN (SELECT DISTINCT UserID FROM vw_search ))
AND UserProfile.PropertyDefinitionID=29 AND
dbo.fnCommonFriendsStep(Users.UserID,36) >=0 ORDER By Degree ASC


fnCommonFriendstep one calculates relationship between the loggedin
member and other members.

if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].
[fnCommonFriendsStep]') and xtype in (N'FN', N'IF', N'TF'))
drop function [dbo].[fnCommonFriendsStep]
GO

SET QUOTED_IDENTIFIER ON
GO
SET ANSI_NULLS ON
GO

CREATE FUNCTION dbo.fnCommonFriendsStep
(
@Want1 INT,
@Want2 INT
)
RETURNS INT
AS

BEGIN
IF @Want1 = @Want2
RETURN 0

DECLARE @Friends TABLE (Generation INT, p INT)
DECLARE @Generation INT

SELECT @Generation = 0

INSERT @Friends
(
Generation,
p
)
SELECT 0,
memberId
FROM network
WHERE friendId = @Want1
UNION
SELECT 0,
friendId
FROM network
WHERE memberId = @Want1

WHILE NOT EXISTS (SELECT 1 FROM @Friends WHERE p = @Want2) AND
@Generation >= 0
BEGIN
SELECT @Generation = @Generation + 1

INSERT @Friends
(
Generation,
p
)
SELECT @Generation,
memberId
FROM network
WHERE friendId IN (SELECT p FROM @Friends WHERE Generation =
@Generation - 1)
AND memberId NOT IN (SELECT p FROM @Friends)
UNION
SELECT @Generation,
friendId
FROM network
WHERE memberId IN (SELECT p FROM @Friends WHERE Generation =
@Generation - 1)
AND friendId NOT IN (SELECT p FROM @Friends)

IF @@ROWCOUNT = 0
SELECT @Generation = -2
END

RETURN @Generation + 1
END
GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS ON
GO


GetRecommendations get the number of recommendations a member has
received.

if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].
[GetRecommendations]') and xtype in (N'FN', N'IF', N'TF'))
drop function [dbo].[GetRecommendations]
GO

SET QUOTED_IDENTIFIER ON
GO
SET ANSI_NULLS ON
GO

CREATE FUNCTION dbo.GetRecommendations
(
@user_id as int
)
RETURNS INT
AS
BEGIN
DECLARE @recomm INT
SELECT @recomm=Count(*) FROM endorsements WHERE memberId=@user_id AND
status='Accepted'
RETURN @recomm
END
GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS ON
GO


GetConnectionsCount fetches number of connections a member has


if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].
[GetConnectionsCount]') and xtype in (N'FN', N'IF', N'TF'))
drop function [dbo].[GetConnectionsCount]
GO

SET QUOTED_IDENTIFIER ON
GO
SET ANSI_NULLS ON
GO

CREATE FUNCTION dbo.GetConnectionsCount
(
@user_id as int
)
RETURNS INT
AS
BEGIN
DECLARE @recCount INT
SELECT @recCount=COUNT(*) FROM network WHERE (memberId=@user_id OR
friendId=@user_id) AND status=1
RETURN @recCount
END
GO
SET QUOTED_IDENTIFIER OFF
GO
SET ANSI_NULLS ON
GO


Please do let me know if you need any more information.