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: Wrong Answer)

#include <iostream>

const int MAXN = 100 + 5,MAXK = 1e5 + 5,MOD = 1e9 + 7;

using namespace std;

int n,k,v[MAXN];
long long dp[MAXN][MAXK],mars[MAXK];

int main()
{
    ///dp[i + 1][j + ceva] += dp[i][j], ceva = [0,a[i + 1]]
    ///j,j + 1,j + 2... j + cva
    ///mars[j] += dp[i][j], mars[j + a[i + 1] + 1] -= dp[i][j]
    cin>>n>>k;
    for(int i = 1; i <= n; i++)
        cin>>v[i];
    for(int j = 0; j <= v[1]; j++)
        dp[1][j] = 1;
    for(int i = 2; i <= n; i++){
        for(int j = 0; j <= k; j++)
            mars[j] = 0;
        for(int j = 0; j <= v[i]; j++){
            mars[j] += dp[i - 1][j];
            mars[j] %= MOD;
            if(j + v[i] + 1 < MAXK){
                mars[j + v[i] + 1] -= dp[i - 1][j];
                mars[j + v[i] + 1] %= MOD;
            }
        }
        for(int j = 0; j <= k; j++){
            mars[j] += mars[j - 1];
            mars[j] %= MOD;
        }
        for(int j = 0; j <= k; j++)
            dp[i][j] = mars[j];
    }
    cout<<dp[n][k] + MOD;
    return 0;
}