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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e9+7;

ll dp[101][100001];
ll cumil[101][100001];
ll n, k;
ll arr[101];
ll pre[101];
int main(){

    cin>>n>>k;

    for(int x=1;x<=n;x++)
        cin>>arr[x];

    for(int x=1;x<=n;x++){
        pre[x] = pre[x-1] + arr[x];
    }

    for(int x=0;x<=k;x++){
        if(x > arr[1]){
            cumil[1][x] = cumil[1][x-1];
            continue;
        }
        if(arr[1] >= x)
            dp[1][x] = 1;
        cumil[1][x] = x+1;
    }

    for(int x=2;x<=n;x++){
        for(ll y = 0; y <= k; y++){
            if(pre[x] < y) break;
            dp[x][y] += cumil[x-1][min(arr[x], y)];
            if(arr[x] < y) dp[x][y] -= cumil[x-1][y - arr[x]-1];
            dp[x][y] %= mod;
            dp[x][y] += mod;
            dp[x][y] %= mod;

            if(y==0) cumil[x][y] = dp[x][y];
            else cumil[x][y] = dp[x][y] + cumil[x][y-1];
            cumil[x][y] %= mod;

//            for(int z = 0; z<= min(arr[x], y); z++){
//                if(y-z<0) break;
//                dp[x][y] += dp[x-1][y - z];
//                dp[x][y] %= mod;
//            }
        }
    }
//    for(int x=1;x<=n;x++){
//        for(int y= 0; y<=5; y++)
//            cout<<cumil[x][y]<<" ";
//        cout<<endl;
//    }
//    cout<<endl;
//    for(int x=1;x<=n;x++){
//        for(int y= 0; y<=5; y++)
//            cout<<dp[x][y]<<" ";
//        cout<<endl;
//    }

    cout<<(dp[n][k] % mod + mod) % mod<<endl;
}