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;
#define rep(i, a) for(int i=0; i<(int)(a); i++)
#define rep1(i, a, b) for(int i=a; i<=(int)(b); i++)
typedef long long LL;

const int MAX_N = 101, MAX_K=1e5+1, mod = 1e9+7;
int n, K, A[MAX_N], dp[MAX_N][MAX_K];

int main() {
    scanf("%d %d", &n, &K);
    rep(i, n) scanf("%d", &A[i]);
    rep(i, n+1) dp[i][0]=1;
    rep1(i, 1, n) {
        rep1(j, 1, K) {
            dp[i][j] = (dp[i][j-1]+dp[i-1][j]) % mod;
            if(j-1-A[i-1]>=0) dp[i][j] = (dp[i][j]-dp[i-1][j-1-A[i-1]]+mod)%mod;
        }
    }
    // rep(i,n+1) rep(j,K+1) {
    //     printf("%d%c", dp[i][j], ((j==K)?'\n':' '));
    // }


    printf("%d\n", dp[n][K]);
    return 0;
}