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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>


using namespace std;

const int MOD = 1E9 + 7;
int main() {
        int n, k;
        scanf("%d%d", &n, &k);
        vector<long long> dp(k + 1);
        dp[0] = 1;
        for (int i = 0, w; i < n; i++) {
                scanf("%d", &w);
                vector<long long> pre(k + 2);
                for (int a = k; a >= 0; a--) {
                        (pre[a + 1] += dp[a]) %= MOD;
                        (pre[a + min(w, k - a) + 1] -= dp[a]) %= MOD;
                }
                for (int j = 0; j <= k; j++) {
                        if (j) (pre[j] += pre[j - 1]) %= MOD;
                        (dp[j] += pre[j]) %= MOD;
                }
        }

        printf("%lld\n", dp[k]);
}