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()))
mod = 1000000007
## dp[i][j] i番目までの子供にj個の飴を分け与える通り数の""jについての累積和""

dp = [[0]*(k+2) for i in range(n)]
for j in range(1,k+2):
  if j-1 <= a[0] : 
    dp[0][j] = dp[0][j-1] + 1
  else:
    dp[0][j] = dp[0][j-1] + 0

for i in range(1,n):
  for j in range(k+2):
      dp[i][j] = (dp[i][j-1] + dp[i-1][j]-dp[i-1][max(j-a[i]-1,0)])%mod

print((dp[n-1][k+1]-dp[n-1][k])%mod)

## 累積和を使わない

#dp = [[0]*(k+1) for i in range(n)]
#
#for j in range(k+1):
#  if j <= a[0] : 
#    dp[0][j] = 1
#  else:
#    dp[0][j] = 0
#for i in range(1,n):
#  for j in range(k+1):
#    for jj in range(max(0,j-a[i]),j+1): 
#      dp[i][j] += dp[i-1][jj]
#print(dp[n-1][k]%1000000007)