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;

typedef long long LL;
const LL MAXN=128;
const LL MAXK=100000+100;
const LL MOD=1e9+7;
LL a[MAXN],dp[MAXN][MAXK],sum[MAXN][MAXK];
LL n,k;

int main(){
	cin>>n>>k;
	for(LL i=1;i<=n;i++){
		cin>>a[i];
	}
	dp[0][0]=1;
	sum[0][0]=1;
	for(LL i=1;i<=k;i++){
		sum[0][i]=1;
	}
	for(LL i=1;i<=n;i++){
		for(LL j=0;j<=k;j++){
			dp[i][j]=(sum[i-1][j]+MOD-(j-a[i]-1<0?0:sum[i-1][j-a[i]-1]))%MOD;
			if(j) sum[i][j]=(sum[i][j-1]+dp[i][j])%MOD;
			else sum[i][j]=dp[i][j];
		}
	}

	cout<<dp[n][k]<<endl;
	return 0;
}
