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 rep(i,n) for(int i=0; i<(n); i++)
#define INF 1000000007
typedef long long ll;
int main(){
	int N,K;
	cin >> N >> K;
	vector<int> a(N);
	rep(i,N) cin >> a[i];

	vector<vector<ll>> dp(N+1,vector<ll>(K+10));
	dp[0][0] = 1;
	for(int i=1; i<=N; i++){
		for(int j=0; j<=K; j++){
			dp.at(i).at(j) = dp.at(i).at(j)+dp.at(i-1).at(j);
			if(j+a.at(i-1) > K) continue;
			dp.at(i).at(j+a.at(i-1)+1) -= dp.at(i-1).at(j);
		}

		for(int j=1; j<=K; j++){
			dp[i][j] = (dp[i][j] + dp[i][j-1]) % INF;
		}
	}

	cout << dp[N][K] << endl;
	return 0;
}