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)

#pragma comment(linker, "/stack:20000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
 
#include <stdio.h>
#include <bits/stdc++.h>
 
using namespace std;
 
#include <x86intrin.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template <typename T> using ordered_set = tree<T, null_type, std::less<T>, rb_tree_tag, tree_order_statistics_node_update>;
  
#define pb push_back
#define mp make_pair
#define in insert
#define f first
#define s second
#define ll long long
#define dd double
  
const int mod = 1e9 + 7;
const int N = 2e5 + 7;
const int inf = 2e9;
const ll INF = 1e18;

ll a[101], dp[200][202020];
ll pre_sum[200][202020];

int main() {
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	// freopen("*.in", "r", stdin);freopen("*.out", "w", stdout);
	ll n, k;
	cin >> n >> k;
	for (int i = 1; i <= n; i++) cin >> a[i];

	for (int i = 0; i <= a[1]; i++) dp[1][i] = 1;
	for (int i = 0; i <= k; i++) {
		if(i == 0) pre_sum[1][i] = dp[1][i] % mod;
		else pre_sum[1][i] = (pre_sum[1][i - 1] + dp[1][i]) % mod;
	}

	for (int i = 2; i <= n; i++) {
		for (int j = 0; j <= k; j++) {
			if(j - min(j * 1ll, a[i])) dp[i][j] = (pre_sum[i - 1][j] - pre_sum[i - 1][j - min(j * 1ll, a[i]) - 1]) % mod;
			else dp[i][j] = pre_sum[i - 1][j] % mod;
		}
		for (int x = 0; x <= k; x++) {
			if(x == 0) pre_sum[i][x] = dp[i][x] % mod;
			else pre_sum[i][x] = (pre_sum[i][x - 1] + dp[i][x]) % mod;
		}
	}

	cout << dp[n][k];
}