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;

const int N = 110,mod=1e9+7;

typedef long long ll;

int a[N];
ll s[110000];
ll f[110][110000];
int main()
{
	ll n,k;
	cin>>n>>k;

	
	memset(f,0,sizeof(f));
	for(int i=1;i<=n;i++) cin>>a[i];
	for(int i =0;i<=k;i++) f[1][i] = (i<=a[1])? 1:0;
	for(int i=2;i<=n;i++)
	{	memset(s,0,sizeof(s));
		s[0]=f[i-1][0];
		for(int j=1;j<=k;j++)
		{
			s[j]=s[j-1]+f[i-1][j];
		}
		for(int j=k;j>=0;j--)
		{
			if(j-a[i]-1>=0) f[i][j]=s[j]-s[j-a[i]-1];
			else f[i][j]=s[j];
			f[i][j]%=mod;
		}
	}
	cout<<f[n][k]<<endl;
 
}