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)

// SPDX-License-Identifier: Apache-2.0
// Copyright (C) 2020 Bjarni Dagur Thor Karason

#include <bits/stdc++.h>

using namespace std;

#define rep(i,a,b) for (__typeof(a) i=(a); i<(b); i++)
#define MP make_pair
#define PB push_back
typedef long long ll;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<int> vi;
#define INF 1000000000
#define PI 3.14159265359

int n, k, mod = 1e9 + 7;
ll memo[100010];

int main()
{
    ios_base::sync_with_stdio(false); cin.tie(nullptr);
    cin >> n >> k;
    memo[0] = 1;
    for (int i = 0; i < n; i++) {
        int a;
        cin >> a;
        vi diff(100010, 0);
        for (int have = k; have >= 0; have--) {
            int l = have + 1, r = have + min(a, k - have);
            if (l <= r) {
                diff[l] += memo[have];
                diff[l] %= mod;
                if (r + 1 <= k) {
                    diff[r + 1] -= memo[have];
                    if (diff[r + 1] < 0) diff[r + 1] += mod;

                }
            }
        }
        int pre = 0;
        for (int i = 0; i <= k; i++) {
            pre += diff[i];
            pre %= mod;
            memo[i] += pre;
            memo[i] %= mod;
        }
    }
    cout << memo[k] << endl;

    return 0;
}