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;
#define ll long long
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)
const int N = 1e5, M = 7 + 1e9;
int n, k, a[101], dp[101][N+1];
/*
	CHEAT MODE
*/
int main()
{
	cin >> n >> k;
	f(i, 1, n+1)
		cin >> a[i];
	dp[0][0] = 1;
	f(i, 1, n+1){
		vector<int> sum(k+1);
		f(w, 0, k+1)
			sum[w] = dp[i-1][w];
		f(w, 1, k+1)
			sum[w] = (sum[w] + sum[w-1])%M; // cumulative sum
		f(w, 0,k+1){ //sum[w-a[i] -> w ] 
			dp[i][w] = (sum[w] - (w-a[i]-1 < 0 ? 0 : sum[w-a[i]-1]) + M)%M;//rang
		}
	}
	cout << dp[n][k];
}