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: Accepted)

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

ll dp[105][100005];
ll sums[100005];
const ll MOD_VAL = 1e9+7;

int n, k;
int a[100005];

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin>>n>>k;
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    for(int i=0;i<=a[0];i++){
        dp[0][i] = 1;
    }
    for(int i=1;i<n;i++){
        for(int j=1;j<=k+1;j++){
            sums[j] = dp[i-1][j-1] + sums[j-1];
            sums[j] %= MOD_VAL;
        }
        for(int j=0;j<=k;j++){
            dp[i][j] += (sums[j+1] - sums[max(j-a[i],0)] + MOD_VAL)%MOD_VAL;
        }
    }
    cout<<dp[n-1][k];

    return 0;
}