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.

Examples

Input

3 4
1 2 3


Output

5


Input

1 10
9


Output

0


Input

2 0
0 0


Output

1


Input

4 100000
100000 100000 100000 100000


Output

665683269
#include <bits/stdc++.h>
using namespace std;

const int mxN=100, mxK=1e5, M=1e9+7;
int n, a[mxN], k;
long long dp[mxN+1][mxK+1];

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

	cin >> n >> k;
	for(int i=0; i<n; ++i)
		cin >> a[i];
	dp[0][0]=1;
	for(int i=0; i<n; ++i) {
		for(int j=1; j<=k; ++j)
			dp[i][j]=(dp[i][j]+dp[i][j-1])%M;
		for(int j=0; j<=k; ++j) {
			dp[i+1][j]=dp[i][j];
			if(j-a[i]-1>=0)
				dp[i+1][j]+=M-dp[i][j-a[i]-1];
			dp[i+1][j]%=M;
		}
	}
	cout << dp[n][k];
}