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)

#include <iostream>
#include <cstring>
#include <algorithm>
#define MOD 1000000007
using namespace std;

long long int n, k, arr[101], dp[101][100001], pre[101][100001], top=0;

int main(void){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>n>>k;
	for(int i=0;i<n;i++){
      	cin>>arr[i];
    	top+=arr[i];
    }
    sort(arr, arr+n);
	for(int i=0;i<=min(k, arr[0]);i++){
		dp[0][i]=1;
		pre[0][i]=i+1;
	}
    int ek=arr[0];
	for(int i=1;i<n;i++){
		dp[i][0]=1;
		pre[i][0]=1;
		for(int j=1;j<=min(arr[i]+ek, k+1);j++){
			if(j-arr[i]-1>-1)
				dp[i][j]=(pre[i-1][min(j, ek)]-pre[i-1][j-arr[i]-1])%MOD;
			else 
				dp[i][j]=pre[i-1][min(j, ek)];
			pre[i][j]=(dp[i][j]+pre[i][j-1])%MOD;
		}
        if(ek+arr[i]<=k)
        	ek+=arr[i];
	}
	cout<<dp[n-1][k];
	return 0;
}