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())
a=list(map(int,input().split()))
#n,k=100,100000
#a=[k//2]*n
mod=pow(10,9)+7
dp=[0]*(k+1) # dp[j]:j個の飴を分ける分け方。人が一人以上のとき、0個の時は明らかに1通り。0人に飴を分ける方法はないので0通り
sdp=[0]*(k+2)
for i in range(a[0]+1):#一人目だけいる時、その人が持てる飴の個数以下なら1
  dp[i]=1 
for i in range(1,n):
      # 累計和にする
      s=0
      for l in range(k+1):
            s+=dp[l]
            sdp[l+1]=s%mod
      for j in range(1,k+1):
        dp[j]=(sdp[1+j]-sdp[max(0,j-a[i])])%mod
print(dp[-1]%mod)