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 = map(int, input().split())
M=10**9+7
As = list(map(int, input().split()))
dp = [[0]*(K+1)for _ in range(N+1)]
dp2 = [[1]*(K+1)for _ in range(N+1)]
dp[0][0] = 1

for i in range(1, N+1):
    for j in range(K+1):
#        print(i,j, dp[i][j-1] - dp2[i][j-1] + dp[i-1][j], j-As[i-1])
        dp[i][j] = 1 if j==0 else (dp[i][j-1] - dp2[i][j-1] + dp[i-1][j])%M
        dp2[i][j] = 0 if j-As[i-1] <0 else dp[i-1][j-As[i-1]]%M
#        dp[i][j] = sum(dp[i-1][max(0, j-As[i])...j])
#        sum(dp[i][0...j]) = sum(dp[i-1][max(0,j-As[i])...j])
#        dp[i][j] = dp[i-1][j] + dp[i-1][j-1] + ... dp[i-1][max(0, j-As[i])]
#print(dp,dp2)
print(dp[N][K])
