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 ll long long int
#define mp make_pair
#define pb push_back
#define ff first 
#define ss second
#define mod 1000000007
#define lim 100005
#define nfs_mw ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;

int main()
{
	nfs_mw;
	int i,j,n,k,v; ll pdp[lim] = {}, pdp2[lim] = {};
	cin>>n>>k;
	pdp[k] = 1;
	for(i=1; i<=n; i++){
		cin>>v;
		pdp2[0] = pdp[v];
		for(j=1; j<=k; j++) pdp2[j] = (((pdp[min(j+v,k)] + (mod - pdp[j-1])%mod)%mod) + pdp2[j-1])%mod;
		swap(pdp,pdp2);
	}
	cout<<pdp[0];
}