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;

const int KMAX = 1e5;
const int mod = 1e9 + 7;

int N, K;
int a[105];
int dp[105][KMAX + 5];
int sum[105][KMAX + 5];

int getsum(int i, int st, int dr)
{
    if(st < 0)  st = 0;
    if(dr > K)  dr = K;
    if(st == 0) return sum[i][dr];
    return (sum[i][dr] - sum[i][st - 1] + mod) % mod;
}

int main()
{
    cin >> N >> K;
    for(int i = 1; i <= N; i++) cin >> a[i];

    dp[0][0] = 1;
    sum[0][0] = 1;
    for(int i = 1; i <= K; i++) sum[0][i] = 1;
    for(int i = 1; i <= N; i++)
        for(int j = 0; j <= K; j++)
        {
            dp[i][j] = getsum(i - 1, j - a[i], j);
            sum[i][j] = dp[i][j];
            if(j > 0)   sum[i][j] = (sum[i][j - 1] + sum[i][j]) % mod;
        }

    cout << dp[N][K] << '\n';
    exit(0);
}