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 PyPy3 (2.4.0) (judge verdict: Accepted)

n, k = map(int, input().split())
a = [int(x) for x in input().split()]
mod = 10**9 + 7

dp = [[0 for _ in range(k+1)] for __ in range(n)]

sm = 0

for i in range(n):
    sm += a[i]
    for j in range(k+1):
        if j == 0:
            dp[i][j] = 1
            continue
        if sm < j:
            dp[i][j] = 0
        else:
            if j-1-a[i] >= 0:
                dp[i][j] = dp[i][j-1] + dp[i-1][j] - dp[i-1][j-1-a[i]]
            else:
                dp[i][j] = dp[i][j-1] + dp[i-1][j]
            dp[i][j] %= mod

print(dp[n-1][k])