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 <bits/stdc++.h>
#define mod 1000000007
using namespace std;

typedef unsigned long long ll;
const int N(105), K(133333);
int a[N], s[K];
ll dp[N][K];
int n, k;

int main()
{
    cin >> n >> k;
    for (int i = 0; i < n; ++i)
        cin >> a[i];
    for (int j = 0; j <= k; ++j)
        dp[0][j] = 0;
    for (int j = 0; j <= n; ++j)
        dp[j][0] = 1;
    for (int i = 1; i <= n; ++i)
    {
        s[0] = 0;
        for (int j = 1; j <= k + 1; j++)
            s[j] = (s[j - 1] + dp[i - 1][j - 1]) % mod;
        for (int j = 1; j <= k; j++)
        {
            dp[i][j] = s[j + 1] - s[max(0, j - a[i - 1])];
            dp[i][j] = dp[i][j] % mod;
            /*
            if (dp[i][j] < 0)
                dp[i][j] += mod;
                */
        }
    }
    cout << dp[n][k];
    return 0;
}