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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <cmath>
#include <iomanip>

using namespace std;

typedef long long ll;

#define INF 1000000009LL
#define MOD 1000000007LL

ll dp[105][100005];

int main() {
  int N, K; cin >> N >> K;
  vector<int> a(N);
  for (int i = 0; i < N; i++) cin >> a[i];
  
  dp[0][0] = 1;
  for (int i = 1; i <= N; i++) {
    for (int j = 0; j <= K; j++) {
      if (j == 0) dp[i][j] = 1;
      else if (j - a[i-1] - 1 >= 0) {
        dp[i][j] = dp[i-1][j] + dp[i][j-1] - dp[i-1][j - a[i-1] - 1];
      } else {
        dp[i][j] = dp[i-1][j] + dp[i][j-1];
      }
      dp[i][j] %= MOD;
    }
  }
  
  cout << dp[N][K] << endl;
  
  return 0;
}