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>
const int N=100005,mu=1000000007;
int n,k,a[N],dp[N],s[N],sum;
int main(){
	scanf("%d%d",&n,&k);
	for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	dp[0]=s[0]=1;
	for (int i=1;i<=a[1];i++) s[i]=1;
	sum=a[1];
	for (int i=1;i<=n;i++){
		for (int j=0;j<=sum;j++) 
			if (j-a[i]>0) dp[j]=s[j]-s[j-a[i]-1];
			else dp[j]=s[j];
		s[0]=dp[0];
		sum+=a[i+1];
		sum=std::min(sum,k);
		for (int j=1;j<=sum;j++) s[j]=((s[j-1]+dp[j])%mu+mu)%mu;
	}
	printf("%d\n",(dp[k]+mu)%mu);
}