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>

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;

ll dp[100005];
ll ndp[100005];
const int MOD = 1000000007;
int n, k;

void solve() {
  cin >> n >> k;
  dp[0] = 1;
  while(n--) {
    int t;
    cin >> t;
    memset(ndp, 0, sizeof(ndp));
    ll curr = 0;
    for(int i = 0; i <= k; i++) {
      curr += dp[i];
      curr %= MOD;
      curr += MOD;
      curr %= MOD;
      ndp[i] += curr;
      ndp[min(k+1, i+t+1)] -= curr;
    }
    memcpy(dp, ndp, sizeof(dp));
  }
  cout << dp[k] << endl;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL); cout.tie(NULL);
  solve();
}