How to `sum( DISTINCT <column> ) OVER ()` using window function?

Question

I have next data:

Here I already calculated total for conf_id. But want also calculate total for whole partition. eg:
Calculate total suma by agreement for each its order (not goods at order which are with slightly different rounding)

How to sum 737.38 and 1238.3? eg. take only one number among group

(I can not sum( item_suma ), because it will return 1975.67. Notice round for conf_suma as intermediate step)

UPD
Full query. Here I want to calculate rounded suma for each group. Then I need to calculate total suma for those groups

SELECT app_period( '2021-02-01', '2021-03-01' );


WITH
target_date AS ( SELECT '2021-02-01'::timestamptz ),
target_order as (
  SELECT
    tstzrange( '2021-01-01', '2021-02-01') as bill_range,
    o.*
  FROM ( SELECT * FROM "order_bt" WHERE sys_period @> sys_time() ) o
  WHERE FALSE
    OR o.agreement_id = 3385 and o.period_id = 10
),
USAGE AS ( SELECT
  ocd.*,


  o.agreement_id                  as agreement_id,
  o.id                            AS order_id,
  
  (dense_rank() over (PARTITION BY o.agreement_id       ORDER BY o.id                     )) as zzzz_id,
  (dense_rank() over (PARTITION BY o.agreement_id, o.id ORDER BY (ocd.ic).consumed_period )) as conf_id,

  
   sum( ocd.item_suma     ) OVER( PARTITION BY (ocd.o).agreement_id                 ) AS agreement_suma2,

 
  (sum( ocd.item_suma )  OVER( PARTITION BY (ocd.o).agreement_id, (ocd.o).id, (ocd.ic).consumed_period )) AS x_suma,
  (sum( ocd.item_cost )  OVER( PARTITION BY (ocd.o).agreement_id, (ocd.o).id, (ocd.ic).consumed_period )) AS x_cost,
  (sum( ocd.item_suma )  OVER( PARTITION BY (ocd.o).agreement_id, (ocd.o).id, (ocd.ic).consumed_period ))::numeric( 10, 2) AS conf_suma,
  (sum( ocd.item_cost )  OVER( PARTITION BY (ocd.o).agreement_id, (ocd.o).id, (ocd.ic).consumed_period ))::numeric( 10, 2) AS conf_cost,
  max((ocd.ic).consumed) OVER( PARTITION BY (ocd.o).agreement_id, (ocd.o).id, (ocd.ic).consumed_period )                   AS consumed,
  (sum( ocd.item_suma )  OVER( PARTITION BY (ocd.o).agreement_id, (ocd.o).id                           )) AS order_suma2
FROM target_order o
LEFT JOIN order_cost_details( o.bill_range ) ocd
  ON (ocd.o).id = o.id  AND  (ocd.ic).consumed_period && o.app_period
)

SELECT 
  *,
  (conf_suma/6) ::numeric( 10, 2 ) as group_nds,
  (SELECT sum(x) from (SELECT  sum( DISTINCT conf_suma )                       AS x FROM usage sub_u WHERE sub_u.agreement_id = usage.agreement_id GROUP BY agreement_id, order_id) t) as total_suma,
  (SELECT sum(x) from (SELECT (sum( DISTINCT conf_suma ) /6)::numeric( 10, 2 ) AS x FROM usage sub_u WHERE sub_u.agreement_id = usage.agreement_id GROUP BY agreement_id, order_id) t) as total_nds
FROM USAGE
WINDOW w AS ( PARTITION BY usage.agreement_id ROWS CURRENT ROW EXCLUDE TIES)
ORDER BY
  order_id,
  conf_id

My old question

Answer 1

better approach dbfiddle:

1. Assign row_number at each order: row_number() over (partition by agreement_id, order_id ) as nrow
2. Take only first suma: filter nrow = 1

with data as (
  select * from (values 
      ( 1, 1, 1, 1.0049 ), (2, 1,1,1.0049), ( 3, 1,1,1.0049 ) ,
      ( 4, 1, 2, 1.0049 ), (5, 1,2,1.0057), 
      ( 6, 2, 1, 1.53 ), ( 7,2,1,2.18), ( 8,2,2,3.48 )
 ) t (id, agreement_id, order_id, suma)
),
intermediate as (select 
 *,
 row_number() over (partition by agreement_id, order_id ) as nrow,
 (sum( suma ) over ( partition by agreement_id, order_id ))::numeric( 10, 2) as order_suma,
from data)

select 
  *,
  sum( order_suma ) filter (where nrow = 1) over (partition by agreement_id)
from intermediate```