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
#define mod 1000000007
using namespace std;

ll n,k;
vector<ll> v(101);

ll solve(){
	ll dp[n+2][k+2];
	
	for(ll i=0;i<=k;i++){
	  dp[1][i]=(i > v[1])?0:1;
	}
	
	for(ll i=2;i<=n;i++){
		for(ll j=0;j<=k;j++){
			if(j==0)
				dp[i][j]=dp[i-1][j]%mod;
			else{
				dp[i][j]=(mod + dp[i][j-1]+dp[i-1][j]-((j-v[i]-1>=0)?dp[i-1][j-v[i]-1]:0))%mod;
			}
		}
	}
	return dp[n][k]%mod;
}

int main(){
	cin>>n>>k;
	for(ll i=1;i<=n;i++){
		cin>>v[i];
	}
	cout<<solve()<<endl;
}