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
readline = sys.stdin.readline

# dp[i][j] = i番目の子供まででj個の飴を配る場合の数
# dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1] + ... + dp[i - 1][j - A[i]]
# 累積和を使える

N,K = map(int,readline().split())
A = list(map(int,readline().split()))
DIV = 10 ** 9 + 7

dp = [[0] * (K + 1) for i in range(N + 1)]
dp[0][0] = 1
sdp = [[0] * (K + 1) for i in range(N + 1)]
for i in range(len(sdp[0])):
  sdp[0][i] = 1

for i in range(1, N + 1):
  for j in range(K + 1):
    # dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1] + ... + dp[i - 1][j - A[i]]
    # 左端をj - A[i], 右端をjとする累積和
    dp[i][j] = sdp[i - 1][j]
    if j - A[i - 1] - 1 >= 0:
      dp[i][j] -= sdp[i - 1][j - A[i - 1] - 1]
    dp[i][j] %= DIV
  sdp[i][0] = dp[i][0]
  for j in range(1, K + 1):
    sdp[i][j] = sdp[i][j - 1] + dp[i][j]
    sdp[i][j] %= DIV

print(dp[N][K])
