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 N=100005,mod=1e9+7;
int dp[2][N],sum[2][N],a[N],n,k;
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int j=0;j<=k;j++)
		sum[0][j]=1;
	for(int i=1;i<=n;i++)
	{
		int p=i&1;
		for(int j=0;j<=k;j++)
			dp[p][j]=(sum[p^1][j]-(j>a[i]?sum[p^1][j-a[i]-1]:0)+mod)%mod;
		for(int j=0;j<=k;j++)
			sum[p][j]=((j?sum[p][j-1]:0)+dp[p][j])%mod;
	}
	printf("%d\n",dp[n&1][k]);
	return 0;
}
