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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdarg.h>
#include <stdio.h>

#define DENOM 1000000007LL

int max(int len, ...) {
    int i, _max, num;
    va_list ap;
    va_start(ap, len);

    _max = va_arg(ap, int);
    for (i = 1; i < len; i++) {
        num = va_arg(ap, int);
        if (num > _max)
            _max = num;
    }
    va_end(ap);
    return _max;
}

int min(int len, ...) {
    int i, _min, num;
    va_list ap;
    va_start(ap, len);

    _min = va_arg(ap, int);
    for (i = 1; i < len; i++) {
        num = va_arg(ap, int);
        if (num < _min)
            _min = num;
    }
    va_end(ap);
    return _min;
}

int main(void) {
    int i, j, l, m, N, K, a[110];
    long long dp[110][100010] = {0LL}, cum[100010]={0LL};
    scanf("%d%d", &N, &K);
    for (i=1;i<=N;i++) scanf("%d", &a[i]);

    for (j=0;j<=a[1];j++) {
        dp[1][j] = 1LL;
    }
    m = a[1];
    for (i=2;i<=N;i++) {
        m += a[i];
        cum[0] = 0LL;
        for (l=1;l<=min(2,m,K)+1;l++) {
            cum[l] = cum[l-1] + dp[i-1][l-1];
            cum[l] %= DENOM;
        }
        for (j=0;j<=min(2,m,K);j++) {
            dp[i][j] = cum[j+1] - cum[((int) max(2,0,j-a[i]))];
            dp[i][j] %= DENOM;
        }
    }

    printf("%lld\n", dp[N][K]);
    return 0;
}