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>
#define MOD 1000000007
using namespace std;
int N,K;

long long int add(long long int a,long long int b){
	return ((a%MOD)+(b%MOD))%MOD;
}

long long int sub(long long int a,long long int b){
	return ((a%MOD)-(b%MOD))%MOD;
}

long long int DP[101][100001], prefix[101][100001];
int A[101];

int solve(){
	for(int i=0;i<=N;i++) DP[i][0]=prefix[i][0]=1;
	for(int i=1;i<=K;i++) prefix[0][i]=1;
	
	for(int i=1;i<=N;i++){
		for(int j=1;j<=K;j++){
			long long int first=0;
			if(j>A[i]){
				first=prefix[i-1][j-A[i]-1];
			}
			long long int last=prefix[i-1][j];
			DP[i][j]=sub(last,first);
			prefix[i][j]=add(DP[i][j],prefix[i][j-1]);
		}
	}
	return DP[N][K];
}

int main(){
	cin>>N>>K;
	for(int i=1;i<=N;i++){
		cin>>A[i];	
	}
	cout<<solve()<<"\n";
	return 0;
}