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)

p = 10**9+7
N,K = map(int,input().split())
A = list(map(int,input().split()))
A.insert(0,0)
dp = [[0 for _ in range(K+1)] for _ in range(N+1)]
dq = [[0 for _ in range(K+2)] for _ in range(N+1)]
for j in range(K+1):
    if j>A[N]:
        dp[N][j] = 0
    else:
        dp[N][j] = 1
    dq[N][j+1] = (dq[N][j]+dp[N][j])%p
for i in range(N-1,0,-1):
    for j in range(K+1):
        dp[i][j] = (dq[i+1][j+1]-dq[i+1][max(0,j-A[i])])%p
        dq[i][j+1] = (dq[i][j]+dp[i][j])%p
print(dp[1][K])