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.

Examples

Input

3 4
1 2 3


Output

5


Input

1 10
9


Output

0


Input

2 0
0 0


Output

1


Input

4 100000
100000 100000 100000 100000


Output

665683269
N, K = list(map(int, input().split()))
A = list(map(int, input().split()))
C = 10**9+7

dp = [[0]*(K+1) for _ in range(N)]
dp[0] = [1]*(A[0]+1)+[0]*(K-A[0])
csum = [[0]*(K+2) for _ in range(N)]

# l=max(0,j[A[i]])のとき
# dp[i][j]=sum(dp[i][l,j+1])としたいがO(NK**2)で間に合わない
# ->和を取る操作を高速化するために累積和のテーブルを準備する

for i in range(1, N):
    for j in range(K+1):
        csum[i-1][j+1] = (csum[i-1][j]+dp[i-1][j]) % C
        l = max(0, j-A[i])
        dp[i][j] = (csum[i-1][j+1]-csum[i-1][l]) % C
print(dp[-1][-1])
