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;

typedef long long ll;
typedef long double ld;
typedef vector<ll> vi;
typedef pair<ll, ll> pii;
typedef vector<pii> vii;
typedef map<ll, ll> mii;
const int MOD = 1e9 + 7;

#define fastIO() \
ios_base::sync_with_stdio(false); cin.tie(nullptr);
#define endl "\n"
#define pb push_back
#define ff first
#define ss second
#define mp make_pair 

int main() {
    int n;
    cin >> n;
    int k;
    cin >> k;
    vi a(n);
    for (auto& i : a)   cin >> i;
    vector<vi> dp(n, vi(k+1, 0));
    //dp[i][j] = no. of ways that [0..i] share j candies
    for (int j = 0; j <= a[0]; ++j) dp[0][j] = 1;
    for (int i = 1; i < n; ++i) {
        dp[i][0] = 1;
        //cout << dp[i][0] << " ";
        for (int j = 1; j <= k; ++j) {
            dp[i][j] = (dp[i][j - 1] + dp[i - 1][j]) % MOD;
            if (j - 1 - a[i] >= 0)  dp[i][j] = (dp[i][j] - dp[i - 1][j - 1 - a[i]] + MOD) % MOD;
            //cout << dp[i][j] << " ";
        }
        //cout << endl;
    }
    cout << dp[n - 1][k] << endl;
    return 0;
}