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
MAX_INT = int(10e2)
MIN_INT = -MAX_INT
mod = 1000000007
sys.setrecursionlimit(1000000)
def IL(): return list(map(int,input().split()))
def SL(): return input().split()
def I(): return int(sys.stdin.readline())
def S(): return input()

N,K = IL()
a = IL()

dp = [[0]*(K+1) for i in range(N+1)] # dp[i人までの子どもが][j個のアメを配る]
Cs = [[0]*(K+1) for i in range(N+1)] # 累積和

aa = a[0]
num = 0
for i in range(K+1):
  if i <= aa:
    dp[1][i] = 1
    num += 1
  Cs[1][i] = num
for i in range(2,N+1):
  num = 0
  for j in range(K+1):
    if j-a[i-1]-1 < 0:
      dp[i][j] = Cs[i-1][j] %mod
    else:
      dp[i][j] = (Cs[i-1][j] - Cs[i-1][j-a[i-1]-1]) %mod
    num += dp[i][j]
    Cs[i][j] = num

print(dp[N][K] %mod)