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 <cassert>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <limits.h>
#include <map>
#include <queue>
#include <set>
#include <string.h>
#include <vector>

using namespace std;
typedef long long ll;

int main() {
  int N, K;
  cin >> N >> K;

  vector<ll> A;
  ll a;
  ll MOD = 1000000007;

  for (int i = 0; i < N; ++i) {
    cin >> a;
    A.push_back(a);
  }

  ll dp[N + 1][K + 1];
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 1;

  for (int i = 0; i < N; ++i) {
    ll sum = 0;
    ll a = A[i];

    for (int k = 0; k <= K; ++k) {
      sum = (sum + dp[i][k]) % MOD;

      int m = k - a - 1;

      if (m >= 0) {
        sum -= dp[i][m];
      }

      dp[i + 1][k] = sum;
    }
  }

  cout << dp[N][K] << endl;

  return 0;
}