Why am I getting Monday as the first day of the week in SQL Server?

Question

I am trying to group records by week, storing the aggregated date as the first day of the week. However, the standard technique I use for rounding off dates does not appear to work correctly with weeks (though it does for days, months, years, quarters and any other timeframe I've applied it to).

Here is the SQL:

select "start_of_week" = dateadd(week, datediff(week, 0, getdate()), 0);

This returns 2011年08月22日 00:00:00.000, which is a Monday, not a Sunday. Selecting @@datefirst returns 7, which is the code for Sunday, so the server is setup correctly in as far as I know.

I can bypass this easily enough by changing the above code to:

select "start_of_week" = dateadd(week, datediff(week, 0, getdate()), -1);

But the fact that I have to make such an exception makes me a little uneasy. Also, apologies if this is a duplicate question. I found some related questions but none that addressed this aspect specifically.

• Martin Smith: (@@DATEFIRST + DATEPART(DW, @SomeDate)) % 7 remains constant irrespective of @@datefirst setting I think. With Monday = 2.

Answer 1

To answer why you're getting a Monday and not a Sunday:

You're adding a number of weeks to the date 0. What is date 0? 1900年01月01日. What was the day on 1900年01月01日? Monday. So in your code you're saying, how many weeks have passed since Monday, January 1, 1900? Let's call that [n]. Ok, now add [n] weeks to Monday, January 1, 1900. You should not be surprised that this ends up being a Monday. DATEADD has no idea that you want to add weeks but only until you get to a Sunday, it's just adding 7 days, then adding 7 more days, ... just like DATEDIFF only recognizes boundaries that have been crossed. For example, these both return 1, even though some folks complain that there should be some sensible logic built in to round up or down:

SELECT DATEDIFF(YEAR, '2010-01-01', '2011-12-31');
SELECT DATEDIFF(YEAR, '2010-12-31', '2011-01-01');

To answer how to get a Sunday:

If you want a Sunday, then pick a base date that's not a Monday but rather a Sunday. For example:

DECLARE @dt DATE = '1905-01-01';
SELECT [start_of_week] = DATEADD(WEEK, DATEDIFF(WEEK, @dt, CURRENT_TIMESTAMP), @dt);

This will not break if you change your DATEFIRST setting (or your code is running for a user with a different setting) - provided that you still want a Sunday regardless of the current setting. If you want those two answers to jive, then you should use a function that does depend on the DATEFIRST setting, e.g.

SELECT DATEADD(DAY, 1-DATEPART(WEEKDAY, CURRENT_TIMESTAMP), CURRENT_TIMESTAMP);

So if you change your DATEFIRST setting to Monday, Tuesday, what have you, the behavior will change. Depending on which behavior you want, you could use one of these functions:

CREATE FUNCTION dbo.StartOfWeek1 -- always a Sunday
(
 @d DATE
)
RETURNS DATE
AS
BEGIN
 RETURN (SELECT DATEADD(WEEK, DATEDIFF(WEEK, '19050101', @d), '19050101'));
END
GO

...or...

CREATE FUNCTION dbo.StartOfWeek2 -- always the DATEFIRST weekday
(
 @d DATE
)
RETURNS DATE
AS
BEGIN
 RETURN (SELECT DATEADD(DAY, 1-DATEPART(WEEKDAY, @d), @d));
END
GO

Now, you have plenty of alternatives, but which one performs best? I'd be surprised if there would be any major differences but I collected all the answers provided so far and ran them through two sets of tests - one cheap and one expensive. I measured client statistics because I don't see I/O or memory playing a part in the performance here (though those may come into play depending on how the function is used). In my tests the results are:

"Cheap" assignment query:

Function - client processing time / wait time on server replies / total exec time
Gandarez - 330/2029/2359 - 0:23.6
me datefirst - 329/2123/2452 - 0:24.5
me Sunday - 357/2158/2515 - 0:25.2
trailmax - 364/2160/2524 - 0:25.2
Curt - 424/2202/2626 - 0:26.3

"Expensive" assignment query:

Function - client processing time / wait time on server replies / total exec time
Curt - 1003/134158/135054 - 2:15
Gandarez - 957/142919/143876 - 2:24
me Sunday - 932/166817/165885 - 2:47
me datefirst - 939/171698/172637 - 2:53
trailmax - 958/173174/174132 - 2:54

I can relay the details of my tests if desired - stopping here as this is already getting quite long-winded. I was a bit surprised to see Curt's come out as the fastest at the high end, given the number of calculations and inline code. Maybe I'll run some more thorough tests and blog about it... if you guys don't have any objections to me publishing your functions elsewhere.

---

If you're lucky enough to be on SQL Server 2022 or better (or Azure SQL Database / MI), you can use two new functions:

• DATETRUNC
• DATE_BUCKET

They are similarly affected by @@DATEFIRST settings, so you have to work around those. Actually DATETRUNC doesn't really offer any benefit unless you know you will always be in SET DATEFIRST 7 - you can make it work like the other workarounds here, but it's not any less complicated to do so. So, let's focus on DATE_BUCKET(). For February 2025, we want the first day of the week to be February 2nd, February 9th, and so on.

DECLARE @d table(d date);
INSERT @d SELECT d = dateadd(DAY, value, '20250201')
 FROM GENERATE_SERIES(0, 10);
-- I want Sunday as week boundary!
-- with default datefirst for 'merika: 7
 
 SELECT
 d,
 wd = DATEPART(WEEKDAY, d),
 wdn = DATENAME(WEEKDAY, d),
 datebuck_wrong = DATE_BUCKET(WEEK, 1, d), -- wrong! based on Jan 1, 1900
 datetrnc_right = DATETRUNC(WEEK,d), -- correct
 oldapproach = DATEADD(DAY, -(DATEPART(WEEKDAY, d) + @@DATEFIRST + 6)%7, d)
 FROM @d;
-- even if someone made it Monday (or any other day!)
SET DATEFIRST 1;
 SELECT
 d,
 wd = DATEPART(WEEKDAY, d),
 wdn = DATENAME(WEEKDAY, d),
 datebuck_wrong = DATE_BUCKET(WEEK, 1, d), -- still wrong
 datetrnc_wrong = DATETRUNC(WEEK,d), -- wrong!
 oldapproach = DATEADD(DAY, -(DATEPART(WEEKDAY, d) + @@DATEFIRST + 6)%7, d)
 FROM @d;

d	wd	wdn	datebuck_wrong	datetrnc_right	oldapproach
2025年02月01日	7	Saturday	2025年01月27日	2025年01月26日	2025年01月26日
2025年02月02日	1	Sunday	2025年01月27日	2025年02月02日	2025年02月02日
2025年02月03日	2	Monday	2025年02月03日	2025年02月02日	2025年02月02日
2025年02月04日	3	Tuesday	2025年02月03日	2025年02月02日	2025年02月02日
2025年02月05日	4	Wednesday	2025年02月03日	2025年02月02日	2025年02月02日
2025年02月06日	5	Thursday	2025年02月03日	2025年02月02日	2025年02月02日
2025年02月07日	6	Friday	2025年02月03日	2025年02月02日	2025年02月02日
2025年02月08日	7	Saturday	2025年02月03日	2025年02月02日	2025年02月02日
2025年02月09日	1	Sunday	2025年02月03日	2025年02月09日	2025年02月09日
2025年02月10日	2	Monday	2025年02月10日	2025年02月09日	2025年02月09日
2025年02月11日	3	Tuesday	2025年02月10日	2025年02月09日	2025年02月09日

d	wd	wdn	datebuck_wrong	datetrnc_wrong	oldapproach
2025年02月01日	6	Saturday	2025年01月27日	2025年01月27日	2025年01月26日
2025年02月02日	7	Sunday	2025年01月27日	2025年01月27日	2025年02月02日
2025年02月03日	1	Monday	2025年02月03日	2025年02月03日	2025年02月02日
2025年02月04日	2	Tuesday	2025年02月03日	2025年02月03日	2025年02月02日
2025年02月05日	3	Wednesday	2025年02月03日	2025年02月03日	2025年02月02日
2025年02月06日	4	Thursday	2025年02月03日	2025年02月03日	2025年02月02日
2025年02月07日	5	Friday	2025年02月03日	2025年02月03日	2025年02月02日
2025年02月08日	6	Saturday	2025年02月03日	2025年02月03日	2025年02月02日
2025年02月09日	7	Sunday	2025年02月03日	2025年02月03日	2025年02月09日
2025年02月10日	1	Monday	2025年02月10日	2025年02月10日	2025年02月09日
2025年02月11日	2	Tuesday	2025年02月10日	2025年02月10日	2025年02月09日

• db<>fiddle

We can coerce DATE_BUCKET() to work as we want by simply providing its optional fourth parameter - origin - any known Sunday, like the trick we used above with 1905年01月01日. December 1st, 2024 was a Sunday, so we can use that:

DECLARE @d table(d date);
DECLARE @origin_date date = '20241201'; -- known Sunday
INSERT @d SELECT d = dateadd(DAY, value, '20250201')
 FROM GENERATE_SERIES(0, 10);
-- I want Sunday as week boundary!
-- with default datefirst for 'merika: 7
 SELECT
 d,
 wd = DATEPART(WEEKDAY, d),
 wdn = DATENAME(WEEKDAY, d),
 datebuck_wrong = DATE_BUCKET(WEEK, 1, d), -- wrong! based on Jan 1, 1900
 /* added: */
 datebuck_origin = DATE_BUCKET(WEEK, 1, d, @origin_date),
 datetrnc_right = DATETRUNC(WEEK,d), -- correct
 oldapproach = DATEADD(DAY, -(DATEPART(WEEKDAY, d) + @@DATEFIRST + 6)%7, d)
 FROM @d;
-- even if someone made it Monday (or any other day!)
SET DATEFIRST 1;
 SELECT
 d,
 wd = DATEPART(WEEKDAY, d),
 wdn = DATENAME(WEEKDAY, d),
 datebuck = DATE_BUCKET(WEEK, 1, d), -- still wrong
 /* added: */
 datebuck_origin = DATE_BUCKET(WEEK, 1, d, @origin_date),
 datetrnc_wrong = DATETRUNC(WEEK,d), -- wrong!
 datefrst = DATEADD(DAY, -(DATEPART(WEEKDAY, d) + @@DATEFIRST + 6)%7, d)
 FROM @d;

d	wd	wdn	datebuck_wrong	datebuck_origin_right	datetrnc_right	oldapproach
2025年02月01日	7	Saturday	2025年01月27日	2025年01月26日	2025年01月26日	2025年01月26日
2025年02月02日	1	Sunday	2025年01月27日	2025年02月02日	2025年02月02日	2025年02月02日
2025年02月03日	2	Monday	2025年02月03日	2025年02月02日	2025年02月02日	2025年02月02日
2025年02月04日	3	Tuesday	2025年02月03日	2025年02月02日	2025年02月02日	2025年02月02日
2025年02月05日	4	Wednesday	2025年02月03日	2025年02月02日	2025年02月02日	2025年02月02日
2025年02月06日	5	Thursday	2025年02月03日	2025年02月02日	2025年02月02日	2025年02月02日
2025年02月07日	6	Friday	2025年02月03日	2025年02月02日	2025年02月02日	2025年02月02日
2025年02月08日	7	Saturday	2025年02月03日	2025年02月02日	2025年02月02日	2025年02月02日
2025年02月09日	1	Sunday	2025年02月03日	2025年02月09日	2025年02月09日	2025年02月09日
2025年02月10日	2	Monday	2025年02月10日	2025年02月09日	2025年02月09日	2025年02月09日
2025年02月11日	3	Tuesday	2025年02月10日	2025年02月09日	2025年02月09日	2025年02月09日

d	wd	wdn	datebuck	datebuck_origin_right	datetrnc	datefrst
2025年02月01日	6	Saturday	2025年01月27日	2025年01月26日	2025年01月27日	2025年01月26日
2025年02月02日	7	Sunday	2025年01月27日	2025年02月02日	2025年01月27日	2025年02月02日
2025年02月03日	1	Monday	2025年02月03日	2025年02月02日	2025年02月03日	2025年02月02日
2025年02月04日	2	Tuesday	2025年02月03日	2025年02月02日	2025年02月03日	2025年02月02日
2025年02月05日	3	Wednesday	2025年02月03日	2025年02月02日	2025年02月03日	2025年02月02日
2025年02月06日	4	Thursday	2025年02月03日	2025年02月02日	2025年02月03日	2025年02月02日
2025年02月07日	5	Friday	2025年02月03日	2025年02月02日	2025年02月03日	2025年02月02日
2025年02月08日	6	Saturday	2025年02月03日	2025年02月02日	2025年02月03日	2025年02月02日
2025年02月09日	7	Sunday	2025年02月03日	2025年02月09日	2025年02月03日	2025年02月09日
2025年02月10日	1	Monday	2025年02月10日	2025年02月09日	2025年02月10日	2025年02月09日
2025年02月11日	2	Tuesday	2025年02月10日	2025年02月09日	2025年02月10日	2025年02月09日

• db<>fiddle

You might say, "Why not just manually hard-code SET DATEFIRST?" Well, that can't be done inside functions, for example, and also you can't just override users' settings who may be relying on the existing setting for other behaviors.

• Baodad: So, if I consider my weeks to start on Sunday and end on Saturday, I can get the last day of the week for any date @d like this: SELECT DATEADD(wk, DATEDIFF(wk, '19041231', @d), '19041231')
• Joel Coehoorn: Old now, and has a lot of votes and listed fairly high for some searches. May be worth adding a note about datetrunc()/date_bucket()
• Aaron Bertrand: @JoelCoehoorn Added some info about those, thanks!

Answer 2

For these that need to get:

Monday = 1 and Sunday = 7:

SELECT 1 + ((5 + DATEPART(dw, GETDATE()) + @@DATEFIRST) % 7);

Sunday = 1 and Saturday = 7:

SELECT 1 + ((6 + DATEPART(dw, GETDATE()) + @@DATEFIRST) % 7);

Above there was a similar example, but thanks to double "%7" it would be much slower.

• Fandango68: This works great too to get the day number from the start of the week being Sun or Mon. Thanks
• vasja: Alternatively select (datediff(dd,5,cal.D_DATE)%7 + 1) and select (datediff(dd,6,cal.D_DATE)%7 + 1)

Answer 3

For those who need the answer at work and creating function is forbidden by your DBA, the following solution will work:

select *,
cast(DATEADD(day, -1*(DATEPART(WEEKDAY, YouDate)-1), YourDate) as DATE) as WeekStart
From.....

This gives the start of that week. Here I assume that Sundays are the start of weeks. If you think that Monday is the start, you should use:

select *,
cast(DATEADD(day, -1*(DATEPART(WEEKDAY, YouDate)-2), YourDate) as DATE) as WeekStart
From.....

Answer 4

Maybe you need this:

SELECT DATEADD(DD, 1 - DATEPART(DW, GETDATE()), GETDATE())

Or

DECLARE @MYDATE DATETIME
SET @MYDATE = '2011-08-23'
SELECT DATEADD(DD, 1 - DATEPART(DW, @MYDATE), @MYDATE)

Function

CREATE FUNCTION [dbo].[GetFirstDayOfWeek]
( @pInputDate DATETIME )
RETURNS DATETIME
BEGIN
SET @pInputDate = CONVERT(VARCHAR(10), @pInputDate, 111)
RETURN DATEADD(DD, 1 - DATEPART(DW, @pInputDate),
 @pInputDate)
END
GO

• Martin Smith: DATEPART(DW is dependant upon @@datefirst
• Quick Joe Smith: I like the simplicity of this one. It seems to run quite well for very large sets of data too.
• Aaron Bertrand: Why not just make the input parameter DATE then you don't have to do any sub-optimal conversions to VARCHAR and back just to strip any accidental time component that's passed in.
• gandarez: The Convert function was used because the returned value doesn't needs Time values.
• Aaron Bertrand: Yes but the point is that converting to a varchar and back again is expensive. If you just have a DATE parameter then you don't care whether time was included... it gets stripped for you.

Answer 5

This works wonderfully for me:

CREATE FUNCTION [dbo].[StartOfWeek]
(
 @INPUTDATE DATETIME
)
RETURNS DATETIME
AS
BEGIN
 -- THIS does not work in function.
 -- SET DATEFIRST 1 -- set monday to be the first day of week.
 DECLARE @DOW INT -- to store day of week
 SET @INPUTDATE = CONVERT(VARCHAR(10), @INPUTDATE, 111)
 SET @DOW = DATEPART(DW, @INPUTDATE)
 -- Magic convertion of monday to 1, tuesday to 2, etc.
 -- irrespect what SQL server thinks about start of the week.
 -- But here we have sunday marked as 0, but we fix this later.
 SET @DOW = (@DOW + @@DATEFIRST - 1) %7
 IF @DOW = 0 SET @DOW = 7 -- fix for sunday
 RETURN DATEADD(DD, 1 - @DOW,@INPUTDATE)
END

• Aaron Bertrand: This seems to return Monday given today's date, not Sunday. The OP already has a function that returns Monday, he wants it to return Sunday. :-)
• trailmax: doh! I should read questions more careful next time. However, my solution can be easily adjusted, if still required. Seems like OP is happy with the accepted answer anyway -)
• Run CMD: This is the correct solution on my machine, since for me : DATEADD(ww, DATEDIFF(ww,0, CONVERT(DATE,'2017年10月8日') ), 0) returns 2017年10月9日!

Answer 6

Googled this script:

create function dbo.F_START_OF_WEEK
(
 @DATE datetime,
 -- Sun = 1, Mon = 2, Tue = 3, Wed = 4
 -- Thu = 5, Fri = 6, Sat = 7
 -- Default to Sunday
 @WEEK_START_DAY int = 1 
)
/*
Find the fisrt date on or before @DATE that matches 
day of week of @WEEK_START_DAY.
*/
returns datetime
as
begin
declare @START_OF_WEEK_DATE datetime
declare @FIRST_BOW datetime
-- Check for valid day of week
if @WEEK_START_DAY between 1 and 7
 begin
 -- Find first day on or after 1753年1月1日 (-53690)
 -- matching day of week of @WEEK_START_DAY
 -- 1753年1月1日 is earliest possible SQL Server date.
 select @FIRST_BOW = convert(datetime,-53690+((@WEEK_START_DAY+5)%7))
 -- Verify beginning of week not before 1753年1月1日
 if @DATE >= @FIRST_BOW
 begin
 select @START_OF_WEEK_DATE = 
 dateadd(dd,(datediff(dd,@FIRST_BOW,@DATE)/7)*7,@FIRST_BOW)
 end
 end
return @START_OF_WEEK_DATE
end
go

http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=47307

Answer 7

CREATE FUNCTION dbo.fnFirstWorkingDayOfTheWeek
(
 @currentDate date
)
RETURNS INT
AS
BEGIN
 -- get DATEFIRST setting
 DECLARE @ds int = @@DATEFIRST 
 -- get week day number under current DATEFIRST setting
 DECLARE @dow int = DATEPART(dw,@currentDate) 
 DECLARE @wd int = 1+(((@dow+@ds) % 7)+5) % 7 -- this is always return Mon as 1,Tue as 2 ... Sun as 7 
 RETURN DATEADD(dd,1-@wd,@currentDate) 
END

• Fandango68: This is the only function that worked for me in SQL Server 2005. Thank you
• Aaron Bertrand: @Fernando68 Can you explain how other solutions didn't work?
• Fandango68: @AaronBertrand sorry don't recall, but I think I was focusing on a quick answer and I tried yours but for some reason it didn't work for me.
• Aaron Bertrand: @Fernando68 Well, that's very helpful. :-\

Answer 8

For the basic (the current week's Sunday)

select cast(dateadd(day,-(datepart(dw,getdate())-1),getdate()) as date)

If previous week:

select cast(dateadd(day,-(datepart(dw,getdate())-1),getdate()) -7 as date)

Internally, we built a function that does it but if you need quick and dirty, this will do it.

Answer 9

I found some of the other answers long-winded or didn't actually work if you wanted Monday as the start of the week.

Sunday

SELECT DATEADD(week, DATEDIFF(week, -1, GETDATE()), -1) AS Sunday;

Monday

SELECT DATEADD(week, DATEDIFF(week, 0, GETDATE() - 1), 0) AS Monday;

Answer 10

Since Julian date 0 is a Monday just add the number of weeks to Sunday which is the day before -1 Eg. select dateadd(wk,datediff(wk,0,getdate()),-1)

• MMJ: That's it bro!! Why always complicate everything using extensive logic, variables, functions...? Since 1/1/1900 was a Monday, every following week will start on a Monday. So if you always get a Monday as a result, but you want a Sunday... just subtract 1 day! That simple! Be happy ;)

Answer 11

Set DateFirst 1;
Select 
 Datepart(wk, TimeByDay) [Week]
 ,Dateadd(d,
 CASE 
 WHEN Datepart(dw, TimeByDay) = 1 then 0
 WHEN Datepart(dw, TimeByDay) = 2 then -1
 WHEN Datepart(dw, TimeByDay) = 3 then -2
 WHEN Datepart(dw, TimeByDay) = 4 then -3
 WHEN Datepart(dw, TimeByDay) = 5 then -4
 WHEN Datepart(dw, TimeByDay) = 6 then -5
 WHEN Datepart(dw, TimeByDay) = 7 then -6
 END
 , TimeByDay) as StartOfWeek
from TimeByDay_Tbl

This is my logic. Set the first of the week to be Monday then calculate what is the day of the week a give day is, then using DateAdd and Case I calculate what the date would have been on the previous Monday of that week.

Answer 12

This is a useful function for me

/* MeRrais 211126 
select [dbo].[SinceWeeks](0,NULL)
select [dbo].[SinceWeeks](5,'2021-08-31')
*/
alter Function [dbo].[SinceWeeks](@Weeks int, @From datetime=NULL)
Returns date
AS
Begin 
 if @From is null 
 set @From=getdate()
 return cast(dateadd(day, -(@Weeks*7+datepart(dw,@From)-1), @From) as date)
END

Answer 13

I don't have any issues with any of the answers given here, however I do think mine is a lot simpler to implement, and understand. I have not run any performance tests on it, but it should be neglegable.

So I derived my answer from the fact that dates are stored in SQL server as integers, (I am talking about the date component only). If you don't believe me, try this SELECT CONVERT(INT, GETDATE()), and vice versa.

Now knowing this, you can do some cool math equations. You might be able to come up with a better one, but here is mine.

/*
TAKEN FROM http://msdn.microsoft.com/en-us/library/ms181598.aspx
First day of the week is
1 -- Monday
2 -- Tuesday
3 -- Wednesday
4 -- Thursday
5 -- Friday
6 -- Saturday
7 (default, U.S. English) -- Sunday
*/
--Offset is required to compensate for the fact that my @@DATEFIRST setting is 7, the default. 
DECLARE @offSet int, @testDate datetime
SELECT @offSet = 1, @testDate = GETDATE()
SELECT CONVERT(DATETIME, CONVERT(INT, @testDate) - (DATEPART(WEEKDAY, @testDate) - @offSet))

• row1: I find that this doesn't work for me. My @@DATEFIRST is also 7, but if your @testDate is the start of the week, then this returns a date that is the day before.

Answer 14

I had a similar problem. Given a date, I wanted to get the date of the Monday of that week.

I used the following logic: Find the day number in the week in the range of 0-6, then subtract that from the originay date.

I used: DATEADD(day,-(DATEPART(weekday,)+5)%7,)

Since DATEPRRT(weekday,) returns 1 = Sundaye ... 7=Saturday, DATEPART(weekday,)+5)%7 returns 0=Monday ... 6=Sunday.

Subtracting this number of days from the original date gives the previous Monday. The same technique could be used for any starting day of the week.

Answer 15

I found this simple and usefull. Works even if first day of week is Sunday or Monday.

DECLARE @BaseDate AS Date

SET @BaseDate = GETDATE()

DECLARE @FisrtDOW AS Date

SELECT @FirstDOW = DATEADD(d,DATEPART(WEEKDAY,@BaseDate) *-1 + 1, @BaseDate)

Answer 16

Maybe I'm over simplifying here, and that may be the case, but this seems to work for me. Haven't ran into any problems with it yet...

CAST('1/1/' + CAST(YEAR(GETDATE()) AS VARCHAR(30)) AS DATETIME) + (DATEPART(wk, YOUR_DATE) * 7 - 7) as 'FirstDayOfWeek'
CAST('1/1/' + CAST(YEAR(GETDATE()) AS VARCHAR(30)) AS DATETIME) + (DATEPART(wk, YOUR_DATE) * 7) as 'LastDayOfWeek'

• Aaron Bertrand: You can get different answers here if you try different settings for SET DATEFIRST.
• Aaron Bertrand: Well, I didn't down vote, but your answer didn't mention DATEFIRST at all (for three and a half years now), and still doesn't. And you should also avoid regional formats like m/d/y, even in scenarios where the m and d are the same.