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
# (10-x) ways: (0,x)-(9-x,9) -> (x, -x)           <- This should never happens when D > 0
# (x) ways:    (10-x,0)-(9,x-1) -> (x-10, 10 - x)

def solve(d, k, outer):
  if k <= 1:
    return 10**k if d == 0 else 0
  # if d >= 10**k:
  #   return 0

  t = (-d)%10 # 10-d%10 doesn't work when d = 10
  d -= t*10**(k-1) - t
  return (10-t-outer)*solve(abs(d//10), k-2, 0)

D = int(input())
result = 0
for k in range(1,20):
  result += solve(D, k, 1)

print(result)