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
input = sys.stdin.readline
n,k = map(int,input().split())
a = tuple(map(int,input().split()))
mod =10**9+7

#dp[i][j]:0~iまでの子がj個分け合う場合の数
dp = [[0]*(k+1) for i in range(n)]
#b[q+1]-b[p] = dp[x][p]+...+dp[x][q]
b = [0]*(k+2)

for i in range(k+1):
    if i <= a[0]:
        dp[0][i] = 1
    b[i+1] = b[i]+dp[0][i] 

for i in range(1,n):
    for j in range(k+1):
        dp[i][j] = (b[j+1]-b[max(0,j-a[i])])%mod
    #bの更新
    b[0] = 0
    for j in range(k+1):
        b[j+1] = (b[j]+dp[i][j])%mod
        dp[i][j] = dp[i][j]%mod
print(dp[-1][-1])