Problem:
Score : 100 points

Problem Statement

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.

Sample Input 1

3 4
1 2 3

Sample Output 1

5

There are five ways for the children to share candies, as follows:

(0, 1, 3)

(0, 2, 2)

(1, 0, 3)

(1, 1, 2)

(1, 2, 1)

Here, in each sequence, the i-th element represents the number of candies that Child i receives.

Sample Input 2

1 10
9

Sample Output 2

0

There may be no ways for the children to share candies.

Sample Input 3

2 0
0 0

Sample Output 3

1

There is one way for the children to share candies, as follows:

(0, 0)

Sample Input 4

4 100000
100000 100000 100000 100000

Sample Output 4

665683269

Be sure to print the answer modulo 10^9 + 7.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

/*for better*/
#include<bits/stdc++.h>
#define fast ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)
#define f(i,a,b) for(i=a;i<b;i++)
#define fr(i,a,b) for(i=a;i>=b;i--)
#define endl '\n'
#define ll long long int
#define ff first
#define ss second
#define pb push_back
#define mod 1000000007
#define M 998244353
using namespace std;

ll dp[100][100005];

int main()
{
    fast; 
    ll n,k,i,j;
    cin>>n>>k;
  	ll a[n+1];
  	f(i,0,n)
  		cin>>a[i];

  	f(j,0,a[0]+1)
  		dp[0][j]=1;
  	f(i,0,n)
  		dp[i][0]=1;

  	f(i,1,n)
  	{
  		f(j,1,k+1)
  		{
  			if(a[i]>=j)
  				dp[i][j]=(dp[i][j-1]%mod + dp[i-1][j]%mod)%mod;
  			else dp[i][j]=(mod+dp[i][j-1]%mod-dp[i-1][j-a[i]-1]%mod)%mod;
  		}
  	}
  	cout<<dp[n-1][k];
}