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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

#define el '\n'
#define ll long long

const int MAXN = 105;
const int MOD = 1e9+7;
int dp[100005]; // byknya cara membagi candy di i
int a[MAXN];

void tambah(int& a, int b){
  a += b;
  if(a>=MOD) a -= MOD;
}

void kurang(int& a, int b){
  a -= b;
  if(a<0) a += MOD;
}

int main(){
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);

  int n, k;
  cin >> n >> k;
  for(int i = 0; i < n; i++){
    cin >> a[i];
  }

  dp[0] = 1;
  for(int i = 0; i < n; i++){
    vector<int> pref(k+1);
    for(int candy = k; candy >= 0; candy--){
      int l = candy+1;
      int r = candy+a[i];
      r = r > k ? k : r;
      tambah(pref[l],dp[candy]);
      kurang(pref[r+1],dp[candy]);
    }

    int sum_now = 0;
    for(int i = 0; i <= k; i++){
      tambah(sum_now,pref[i]);
      tambah(dp[i],sum_now);
    }
  }
  cout << dp[k] << el;

  return 0;
}