Problem:
Score : 100 points

Problem Statement

There are N children, numbered 1, 2, \ldots, N.

They have decided to share K candies among themselves.
Here, for each i (1 \leq i \leq N), Child i must receive between 0 and a_i candies (inclusive).
Also, no candies should be left over.

Find the number of ways for them to share candies, modulo 10^9 + 7.
Here, two ways are said to be different when there exists a child who receives a different number of candies.

Constraints

All values in input are integers.

1 \leq N \leq 100

0 \leq K \leq 10^5

0 \leq a_i \leq K

Input

Input is given from Standard Input in the following format:

N K
a_1 a_2 \ldots a_N

Output

Print the number of ways for the children to share candies, modulo 10^9 + 7.

Sample Input 1

3 4
1 2 3

Sample Output 1

5

There are five ways for the children to share candies, as follows:

(0, 1, 3)

(0, 2, 2)

(1, 0, 3)

(1, 1, 2)

(1, 2, 1)

Here, in each sequence, the i-th element represents the number of candies that Child i receives.

Sample Input 2

1 10
9

Sample Output 2

0

There may be no ways for the children to share candies.

Sample Input 3

2 0
0 0

Sample Output 3

1

There is one way for the children to share candies, as follows:

(0, 0)

Sample Input 4

4 100000
100000 100000 100000 100000

Sample Output 4

665683269

Be sure to print the answer modulo 10^9 + 7.

Solution in Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

{-# LANGUAGE BangPatterns, MultiWayIf #-}
import Control.Arrow
import Control.Monad.Fix
import Data.Int
import Data.Foldable
import qualified Data.Vector as V
import qualified Data.Vector.Unboxed as VU
import qualified Data.ByteString.Char8 as B
import Debug.Trace

solve :: Int -> Int -> VU.Vector Int32 -> Int32
solve !n !k !an = VU.last $ V.last dp
 where
  mod' x y | x >= 2 * y = x - 2 * y
           | x >= y     = x - y
           | otherwise  = x

  dp :: V.Vector (VU.Vector Int32)
  dp = V.postscanl'
    (\(!dpi) (!i) -> VU.scanl'
      (\(!dpij) (!j) ->
        let !p = if j - fromIntegral (an VU.! i) - 1 >= 0
              then dpi VU.! (j - fromIntegral (an VU.! i) - 1)
              else 0
            !q = dpi VU.! j
            !a = dpij + (q - p)
        in  a `mod'` (10 ^ 9 + 7)
      )
      1
      (VU.generate k (+ 1))
    )
    (VU.generate (k + 1) (\i -> if i == 0 then 1 else 0))
    (V.convert $ VU.generate n id)

main = do
  let readInt :: Integral a => B.ByteString -> Maybe (a, B.ByteString)
      readInt = fmap (first fromIntegral . second B.tail) . B.readInt

  (n, k) <-
    (\(!vec) -> (vec VU.! 0, vec VU.! 1)) . VU.unfoldrN 2 readInt <$> B.getLine
  an <- VU.unfoldrN n readInt <$> B.getLine
  print $ solve n k an