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 ll long long
#define pb push_back
#define mp make_pair
const int N=3050;
const int M=100050;
const int mod=1e9+7;
void add(int &a, int b)
{ a+=b;if(a>=mod) a-=mod;}
void sub(int &a, int b)
{ a-=b;if(a<0) a+=mod;}
int a[N];
int dp[M];
int main()
{
int n,k;
scanf("%i %i",&n,&k);
for(int i=1;i<=n;i++)
scanf("%i",&a[i]),a[i]++;
dp[0]=1;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=k;j++)
add(dp[j],dp[j-1]);
for(int j=k;j>=0;j--)
if(j>=a[i])
sub(dp[j],dp[j-a[i]]);
}
printf("%i",dp[k]);
return 0;
}