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
import sys
MOD=10**9+7
def solve(N:int,K:int,A:list)->int:
    #dp[i][j]:=i人目までの子供にj個の飴を配る場合の数
    dp=[[0]*(K+1) for i in range(N+1)]
    for i in range(N):dp[i][0]=1
    sum=[0]*(K+2)
    for i in range(N):
        for j in range(K+1):#累積和
            sum[j+1]=(sum[j]+dp[i][j])%MOD
        for j in range(K+1):
            dp[i+1][j]=(sum[j+1]-sum[max(0,j-A[i])]+MOD)%MOD
            dp[i+1][j]%=MOD
    return dp[N][K]
def main(lines):
    N,K=map(int,lines[0].split())
    A=list(map(int,lines[1].split()))
    print(solve(N,K,A))
if __name__ == '__main__':
    lines = []
    for l in sys.stdin:
        lines.append(l.rstrip('\r\n'))
    main(lines)
