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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

def f(n, k, a_s):
    md = 10 ** 9 + 7
    dp = [[0] * (k + 1) for _ in range(n)]
    dp[0][0] = 1
    dpi = dp[0]
    for i, a in enumerate(a_s[:-1]):
        dpi1 = dp[i + 1]
        dpi1[0] = s = dp[i][0] % md
        for j in range(1, a + 1):
            dpi1[j] = s = (s + dpi[j]) % md
        for j in range(a + 1, k + 1):
            dpi1[j] = s = (s + dpi[j] - dpi[j - a - 1]) % md
        dpi = dpi1
    return sum(dp[-1][k - a_s[-1]:]) % md


n, k = map(int, input().split())
a_s = list(map(int, input().split()))
print(f(n, k, a_s))