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++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <vector>
#include <utility>
#include <queue>
#include <stack>
#include <cstring>
#include <set>
#include <map>
#include <algorithm>  
#include <cmath>
#include <iomanip>
#define ll long long
using namespace std;
ll mod=pow(10,9)+7;
ll dp[105][100005];
ll n,k;
ll arr[105];
ll calc(ll idx,ll candy)
{
    if(idx==n && candy==k)
    {
        return 1;
    }
    if(idx==n)
    {
        return 0;
    }
    if(candy>k)
    {
        return 0;
    }
    if(dp[idx][candy]!=-1)
    {
        return dp[idx][candy];
    }
    ll total=arr[idx];
    //cout<<total<<endl;
    ll ans1=calc(idx+1,candy);
    for(ll i=1;i<=total;++i)
    {
        ll ans2=0;
        if(candy+i<=k)
        {
            ans2=calc(idx+1,candy+i);
        }
        ans2 %= mod;
        ans1=((ans1%mod)+(ans2%mod))%mod;
    }
    dp[idx][candy]=ans1%mod;
    return dp[idx][candy];
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    memset(dp,-1,sizeof(dp));
    cin>>n>>k;
    for(ll i=0;i<n;++i)
    {
        cin>>arr[i];
    }
    cout<<calc(0,0)<<endl;
}