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;

int n, k;
int a[111];
int dp[111][100100];
int md = 1e9+7;
void addMd(int &a, int b)
{
    if (b < 0) b += md;
    if (b >= md) b -= md;
    a += b;
    if (a >= md) a -= md;
}
int main()
{
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; ++i)
        scanf("%d", &a[i]);
    for (int i = 0; i < n; ++i)
    {
        if (!i)
        {
            for (int j = 0; j <= a[i]; ++j)
                dp[i][j] = 1 + (j ? dp[i][j-1] : 0);
            for (int j = a[i]+1; j <= k; ++j)
                dp[i][j] = dp[i][j-1];
            continue;
        }
        for (int j = 0; j <= k; ++j)
        {
            int lft = max(0, j-a[i]);
            addMd(dp[i][j], dp[i-1][j] - (lft == 0? 0 : dp[i-1][lft-1]));
        }
        for (int j = 1; j <= k; ++j)
            addMd(dp[i][j], dp[i][j-1]);
    }
    addMd(dp[n-1][k], -(k ? dp[n-1][k-1] : 0));
    printf("%d\n", dp[n-1][k]);

    return 0;
}