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<cstdio>
#define maxn 100050
#define HR 1000000007
using namespace std;
int s[maxn],dp[maxn];
int main(){
	int n,k;
	scanf("%d%d",&n,&k);
	s[0]=dp[0]=1;
	for (int i=1;i<=n;i++){
		int x;
		scanf("%d",&x);
		for (int j=1;j<=k;j++)
		s[j]=(s[j-1]+dp[j])%HR;
		for (int j=1;j<=k;j++)
		if (j-x>0) dp[j]=(s[j]-s[j-x-1]+HR)%HR;else
			dp[j]=s[j];
	}
	printf("%d\n",dp[k]);
	return 0;
}

/*
dp[i][j]表示前i个人 一共分了j个的方案数
dp[i][j]=dp[i-1][j-k] (0<=k<=a[i]) 
前缀和优化 
*/