Get the SUM of a DATEDIFF result - sql-server

Question

I have a SQL statement (MS SQL Server 2012) that makes a simple calculation of the differences in dates in some records. I want to return the total / sum of the DATEDIFFs as well.
This is my query:
with cte as (
select ROW_NUMBER () OVER (ORDER BY Date) as ID, Die_ID, Date, Status
from Tooling_Status
where (date between '2018-02-27 00:00:00' and '2019-02-27 11:59:59')
and Date is not null)
select DATEDIFF (ss, c1.Date, min (c2.Date)) as Seconds, convert (nvarchar (10), c1.Date, 120) as DIA
from cte c1
left outer join cte c2
on c1.Date < c2.Date and c1.Die_ID = c2.Die_ID
where c1.Status = 2 and c2.Status = 1
group by c1.Date
order by DIA desc
And this my result:
Seconds DAY
2253 2019-02-27
166 2019-02-27
17 2019-02-27
104 2019-02-27
36 2019-02-27
11 2019-02-07
20 2019-02-07
32 2019-02-07
22 2019-02-07
27 2019-02-07
33 2019-02-07
15 2019-02-07
34 2019-02-07
120 2019-02-04
99420 2019-01-07
60 2018-09-26
I need this result:
Seconds DAY
2732 2019-02-27
194 2019-02-07
120 2019-02-04
99420 2019-01-07
60 2018-09-26
Thanks For Help!

Answer 1

Few adjustments
Extra GROUP BY using a derived table
Removed unnecessary casts to a string
Formatting...
SQL:
WITH cte
AS (SELECT ROW_NUMBER() OVER (ORDER BY Date) AS ID,
Die_ID,
Date,
Status
FROM Tooling_Status
WHERE (date BETWEEN '2018-02-27 00:00:00' AND '2019-02-27 11:59:59' )
AND Date IS NOT NULL
)
SELECT SUM(d.Seconds) AS Seconds
, d.DIA AS [Day]
FROM (
SELECT DATEDIFF(ss, c1.Date, MIN(c2.Date)) AS Seconds,
CAST(c1.Date AS DATE) AS DIA
FROM cte c1
LEFT OUTER JOIN cte c2
ON c1.Date < c2.Date
AND c1.Die_ID = c2.Die_ID
WHERE c1.Status = 2 AND c2.Status = 1
GROUP BY c1.Date
) d
GROUP BY d.DIA
ORDER BY [Day] DESC;