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 C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#pragma GCC optimize "trapv"
#include<bits/stdc++.h>
#define faster ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
using namespace std;
int MOD=1e9+7;

signed main()
{
   faster;
   #ifndef ONLINE_JUDGE
     freopen("ip.txt", "r", stdin);
     freopen("op.txt", "w", stdout);
   #endif
    // int t;cin>>t;
     //while(t--)
     {
          int n,k;
          cin>>n>>k;
          int a[n+1];
          for (int i=1;i<=n;i++)
               cin>>a[i];
          long long  dp[n+1][k+1];
          for (int i=0;i<=n;i++)
               for (int j=0;j<=k;j++)
                    dp[i][j]=0;
          for (int i=0;i<=a[1];i++)
               dp[1][i]=1;
          for (int i=2;i<=n;i++)
               for (int j=0;j<=k;j++)
               {
                    int k=0;
                    while (j-k>=0 && k<=a[i])
                    {
                         dp[i][j]=(dp[i][j]+dp[i-1][j-k])%MOD;
                         k++;
                    }
               }
          cout<<dp[n][k];
     }
}