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>
#define M 1000000007
#define fr(i,a,b) for(i=a;i<=b;i++)
using namespace std;

int main() {
	int n,k,i,j,last,first;
cin>>n>>k;
int a[n+1];
int prefix[n+1][k+1],dp[n+1][k+1];
fr(i,1,n)cin>>a[i];
fr(i,0,n){dp[i][0]=prefix[i][0]=1;}
fr(i,1,k){dp[0][i]=0;}
fr(i,1,k){prefix[0][i]=1;}
fr(i,1,n){
	 fr(j,1,k){
	 	last=j-a[i]-1>=0?prefix[i-1][j-a[i]-1]:0;
	 	first=prefix[i-1][j];
	 	dp[i][j]=(first-last+M)%M;
	 	prefix[i][j]=(prefix[i][j-1]+dp[i][j])%M;
	 }
}
cout<<dp[n][k]<<endl;
	return 0;
}