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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
ll mod=1e9+7;
const int N=1e5+2;
ll dp[N][102];
ll a[106];
ll solve(ll i,ll n,ll k)
{
  if(k==0)
    return 1;
  if(i>n)
    return 0;
  if(dp[k][i]!=-1)
    return dp[k][i];
  ll ans=0;
  for(int j=0;j<=a[i];j++)
    if(k>=j)
    ans=(ans+solve(i+1,n,k-j))%mod;
    return dp[k][i]=ans;
}
int main()
{
  int n,k;
  cin>>n>>k;
  memset(dp,-1,sizeof(dp));
  for(int i=1;i<=n;i++)
    cin>>a[i];
    cout<<solve(1,n,k);
}