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 "bits/stdc++.h"
using namespace std;

#define fi first
#define se second
#define pb push_back
#define int long long
#define all(x) (x).begin(), (x).end()

using LL = long long;
using LD = long double;
using pii = pair<int, int>;
using vii = vector<pii>;

const int INF = 1e18;
const int MOD = 1e9 + 7;
const int N = 1e5 + 5;

// int solve(int n, int k) {
//     if (csum[n] < k) {
//         return 0;
//     }

//     if (n == 1 or csum[n] == k) {
//         return 1;
//     }

//     if (dp[n][k] != -1) {
//         return dp[n][k];
//     }

//     int ans = 0;
//     for (int x = 0; x <= min(a[n], k); x++) {
//         ans = (ans + solve(n - 1, k - x)) % MOD;
//     }

//     return dp[n][k] = ans;
// }

int32_t main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);

    int n, k;
    cin >> n >> k;

    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }

    vector<vector<int>> dp(n + 1, vector<int>(k + 1));
    for (int i = 0; i <= n; i++) {
        dp[i][0] = 1;
    }

    for (int i = 1; i <= n; i++) {
        vector<int> csum(dp[i - 1]);
        for (int j = 1; j <= k; j++) {
            csum[j] = (csum[j] + csum[j - 1]) % MOD;
        }
        for (int j = 1; j <= k; j++) {
            int l = max(0LL, j - a[i]), r = j;
            dp[i][j] = (csum[r] - (l ? csum[l - 1] : 0)) % MOD;
            dp[i][j] = (dp[i][j] + MOD) % MOD;
        }
    }

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