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 <stdio.h>
#include <string.h>

const int MAXN = 105;
const int MAX_K = (int)1e5 + 5;
const int MOD = (int)1e9 + 7;

int a[MAXN];
int dp[MAXN][MAX_K];

int main(void) {
    int n, k;

    scanf(" %d %d", &n, &k);
    for (int i = 1; i <= n; i++) {
        scanf(" %d", &a[i]);
    }

    for (int j = 0; j <= a[n]; j++) {
        dp[n][j] = 1;
    }
    for (int i = n - 1; i >= 0; i--) {
        int idx = 0;
        int sum = 0;
        for (int j = 0; j <= k; j++) {
            sum += dp[i + 1][j];
            sum %= MOD;
            while(j - a[i] > idx) {
                sum -= dp[i + 1][idx];
                sum = (sum % MOD + MOD) % MOD;
                idx++;
            }
            dp[i][j] = sum;
        }
    }
    printf("%d\n", dp[1][k]);

    return 0;
}