For a positive integer n, we denote the integer obtained by reversing the decimal notation of n (without leading zeroes) by rev(n). For example, rev(123) = 321 and rev(4000) = 4.

You are given a positive integer D. How many positive integers N satisfy rev(N) = N + D?

Constraints

* D is an integer.
* 1 ≤ D < 10^9

Input

Input is given from Standard Input in the following format:


D


Output

Print the number of the positive integers N such that rev(N) = N + D.

Examples

Input

63


Output

2


Input

75


Output

0


Input

864197532


Output

1920
def solve(d, K):
  r = 1
  for k in range(K,1,-2):
    if d >= 10**k:
      return 0
    t = (-d)%10
    d = abs((d-t*(10**(k-1)-1))//10)
    r *= 10-t-(K==k)
  return r*(10**(K%2)) if d == 0 else 0

D = int(input())
result = 0
l = len(str(D))
for k in range(l,2*l+1):
  result += solve(D, k)

print(result)