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;
 
#define ll long long
 ll n,k;
 ll dp[109][100010];// (i,x)n of ways to have distribute x candies for the first i kids
 ll dp2[100][100010];// prefix sum of dp  dp2[i][x] sum of dp [i][0,1,2,...x]
 ll a[109];
 
 const ll mod=1e7;
 
 ll sum(int i,int left,int right){
     if(left!=0){
     return dp2[i][right]-dp2[i][left-1];}
      return dp2[i][right];
 }
 
  
int main()
{
    cin>>n>>k;
   
    for(ll i=1;i<=n;i++){
        cin>>a[i];
    }
   
 
//cout<<sum(1,0,1);return 0;
     for(ll i =1;i<=n;i++){
       for(ll j=0;j<=k;j++){
           //j=total of candies distributed
           // if i want to have 10 in total and ai = 5 then i need at least 5 so i nueed sum from 5 upto 10
           
           if(i==1){
              if(j<=a[i]){
               dp[i][j]=1;}
           }else if(j-a[i]>=0){
               dp[i][j]=(dp2[i-1][j]-dp2[i-1][j-a[i]-1])%mod;
           }else{
                dp[i][j]=(dp2[i-1][j])%mod;
           }if(j==0){
               dp2[i][j]=1;
           }else{
               dp2[i][j]=(dp2[i][j-1]+dp[i][j])%mod;
           }
          // cout<<dp[i][j]<<" "<<i<<" "<<j<<endl;
           
       }
           
       }
        //cout<<ans;
   cout<<dp[n][k]%mod;
}