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: Accepted)

#include <iostream>
#include <iomanip>
#include <unordered_map>
#include <set>
#include <fstream>
#include <cstring>
#include <climits>
#include <algorithm>
#include <math.h>
#include <vector>
using namespace std;

#define ll long long
#define mod 1000000007

pair<ll, ll> dp[101][100001];

//dp[i][j].first is the no of ways to distribute j candies among students i onwards
//dp[i][j].second is the no of ways to distribute atmost j candies among students i onwards

ll solve_ite(int* a, int n, int total_candies) {

	for (int i = n - 1; i >= 0; i--) {
		for (int j = 0; j <= total_candies; j++) {
			if (j > a[i])
				dp[i][j].first = dp[i + 1][j].second - dp[i + 1][j - a[i] - 1].second + mod;
			else
				dp[i][j].first = dp[i + 1][j].second;
			if (j == 0)
				dp[i][j].second = dp[i][j].first;
			else
				dp[i][j].second = dp[i][j].first + dp[i][j - 1].second;
			dp[i][j].first %= mod;
			dp[i][j].second %= mod;
		}
	}
	return dp[0][total_candies].first;
}

int main() {
  
	int n, k;
	cin >> n >> k;
	int a[n];
	for (int i = 0; i < n; ++i)
		cin >> a[i];
	for (int i = 0; i < 101; ++i) {
		for (int j = 0; j < 100001; ++j) {
			dp[i][j].first = 0;
			dp[i][j].second = 0;
		}
	}
	dp[n][0].first = 1;
	dp[n][0].second = 1;
	for (int j = 0; j <= k; ++j) {
		dp[n][j].second = 1;
	}
	cout << solve_ite(a, n, k);

	return 0;
}