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;
#define MOD 1000000007
using ll = long long ;
 
ll arr[105],dp[105][100005];
int main(){
	ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	ll i,j,n,k;
	cin>>n>>k;
	for(i=1;i<=n;i++) cin>>arr[i];
	dp[0][0]=1; 
	for(i=1;i<=k;i++) dp[0][i]=0; 
	for(i=1;i<=n;i++){
		for(j=1;j<=k;j++)
			dp[i-1][j]=(dp[i-1][j]+dp[i-1][j-1])%MOD;
			
		for(j=0;j<=k;j++){
			ll mn=j-arr[i]-1;
			dp[i][j]=dp[i-1][j];
			if(mn>=0) dp[i][j]=(dp[i][j]-dp[i-1][mn]+MOD)%MOD;
		}
	}
	cout<<dp[n][k]<<endl;
	return 0;
}