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++ (Clang 10.0.0) (judge verdict: Accepted)

#pragma region
#include <algorithm>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <stack>
#include <string>
#include <vector>
typedef long long ll;
#define rep(i, a, n) for (long long i = a; i <= n; ++i)
#define per(i, a, n) for (long long i = n; i >= a; --i)
#define IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
using namespace std;
/*
 *                                                     __----~~~~~~~~~~~------___
 *                                    .  .   ~~//====......          __--~ ~~
 *                    -.            \_|//     |||\\  ~~~~~~::::... /~
 *                 ___-==_       _-~o~  \/    |||  \\            _/~~-
 *         __---~~~.==~||\=_    -_--~/_-~|-   |\\   \\        _/~
 *     _-~~     .=~    |  \\-_    '-~7  /-   /  ||    \      /
 *   .~       .~       |   \\ -_    /  /-   /   ||      \   /
 *  /  ____  /         |     \\ ~-_/  /|- _/   .||       \ /
 *  |~~    ~~|--~~~~--_ \     ~==-/   | \~--===~~        .\
 *           '         ~-|      /|    |-~\~~       __--~~
 *                       |-~~-_/ |    |   ~\_   _-~            /\
 *                            /  \     \__   \/~                \__
 *                        _--~ _/ | .-~~____--~-/                  ~~==.
 *                       ((->/~   '.|||' -_|    ~~-/ ,              . _||
 *                                  -_     ~\      ~~---l__i__i__i--~~_/
 *                                  _-~-__   ~)  \--______________--~~
 *                                //.-~~~-~_--~- |-------~~~~~~~~
 *                                       //.-~~~--\
 *                       ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
 */
#pragma endregion
const int maxn = 105;
const ll mod = 1e9 + 7;
ll a[maxn], b[int(1e5 + 5)], n, k;
ll dp[int(1e5 + 5)];
int main() {
    IO;
    cin >> n >> k;
    rep(i, 1, n) cin >> a[i];
    dp[0] = 1;
    rep(i, 1, n) {
        memset(b, 0, sizeof(b));
        per(j, 0, k) {
            int l = j + 1, r = min(k, j + a[i]);
            if (l <= r) {
                (b[l] += dp[j]) %= mod;
                (b[r + 1] += mod - dp[j]) %= mod;
            }
        }
        ll now = 0;
        rep(i, 0, k) {
            (now += b[i]) %= mod;
            (dp[i] += now) %= mod;
        }
    }
    cout << dp[k] << endl;
}