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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

'use strict';
(function(input){
  input=input.trim().split(/\s+/).map(x=>x-0);
  const N=input.shift();
  const K=input.shift();
  const M=Math.pow(10,9)+7;
  const a=input;
  const dp=new Array(N).fill(0).map(x=>new Array(K+1).fill(0));
  //dp[i][k]で添字0~iの子供で飴玉残りk個まで消費する場合の数
  dp[-1]=new Array(K+1).fill(0);
  dp[-1][K]=1;
  for(let i=0;i<N;i++){
    for(let k=K;k>=0;k--){
      for(let j=0;j<=a[i]&&k-j>=0;j++){
        dp[i][k-j]+=dp[i-1][k];
        dp[i][k-j]%=M;
      }
    }
  }
  console.log(dp);
  console.log(dp[N-1][0]);
})(require('fs').readFileSync('/dev/stdin','utf8'));