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"
#define all(x) (x).begin(), (x).end()

using namespace std;

const int MSIZE = 100200;

const long long mod = 1e9 + 7;

long long dp[2][MSIZE];
long long pref[MSIZE];

long long Get(int l, int r) {
    l = max(0, l);
    return (mod + pref[r] - (l ? pref[l - 1] : 0LL)) % mod;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    memset(dp, 0, sizeof dp);
    int n, k;
    cin >> n >> k;
    dp[0][0] = 1;
    long long answer = 1;
    
    int * arr = new int[n + 1];
    for (int i = 1; i <= n; i++) {
        cin >> arr[i];
    }

    for (int i = 1; i <= n; i++) {
        memcpy(pref, dp[(i & 1) ^ 1], sizeof pref);
        for (int u = 1; u <= k; u++) {
            pref[u] += pref[u - 1];
            pref[u] %= mod;
        }
        for (int u = 0; u <= k; u++) {
            dp[i & 1][u] = Get(u - arr[i], u);       
        }
    }
    cout << dp[n & 1][k] << endl;
}