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 mod 1000000007
using namespace std;
int n,k;
int a[105];
int dp[105][100005];
int s[105][100005];

int main(){
	scanf("%d %d",&n,&k);
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	for(int i=0;i<=k;i++) s[0][i]=1;
	for(int i=1;i<=n;i++){
		for(int j=0;j<=k;j++){
			if(j<=a[i]) dp[i][j]=s[i-1][j];
			else{
				dp[i][j]=s[i-1][j]-s[i-1][j-a[i]-1];
				if(dp[i][j]<0) dp[i][j]+=mod;
			}
		}
		s[i][0]=dp[i][0];
		for(int j=1;j<=k;j++){
			s[i][j]=s[i][j-1]+dp[i][j];
			s[i][j]%=mod;
		}
	}
	printf("%d\n",dp[n][k]);
	
	return 0;
}