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;

int n, k, a[100010];
long long dp[110][100010], mod = 1e9+7;

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> n >> k;
	for(int i=0; i<n; ++i) cin >> a[i+1];
	dp[0][0] = 1;
	for(int i=1; i<=n; ++i){
		for(int j=0; j<=k; ++j){
			dp[i][j] = dp[i-1][j-min(a[i], j)] - dp[i-1][j+1];
			dp[i][j] %= mod;
			if(dp[i][j] < 0) dp[i][j] += mod;
		}
		for(int j=k-1; j>=0; --j){
			dp[i][j] += dp[i][j+1];
			dp[i][j] %= mod;
		}
	}
	cout << dp[n][k] << "\n";
}