#include <cstdio>
#include <cstdlib>

typedef long long i64;

i64 mem[20];

i64 dfs(int l, int r, i64 d) {
  if (l >= r) return d ? 0 : (l == r ? 10 : 1);
  int t = (10 - d % 10) % 10;
  i64 v = d - t * mem[r - l] + t;
  return ((l ? 10 : 9) - t) * dfs(l + 1, r - 1, llabs(v) / 10);
}

int main() {
  mem[0] = 1;
  for (int i = 1; i <= 18; ++i) mem[i] = 10 * mem[i - 1];
  int d;
  scanf("%d", &d);
  i64 ans = 0;
  for (int i = 0; i <= 18; ++i) ans += dfs(0, i, d);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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

### Response

```cpp
#include <cstdio>
#include <cstdlib>

typedef long long i64;

i64 mem[20];

i64 dfs(int l, int r, i64 d) {
  if (l >= r) return d ? 0 : (l == r ? 10 : 1);
  int t = (10 - d % 10) % 10;
  i64 v = d - t * mem[r - l] + t;
  return ((l ? 10 : 9) - t) * dfs(l + 1, r - 1, llabs(v) / 10);
}

int main() {
  mem[0] = 1;
  for (int i = 1; i <= 18; ++i) mem[i] = 10 * mem[i - 1];
  int d;
  scanf("%d", &d);
  i64 ans = 0;
  for (int i = 0; i <= 18; ++i) ans += dfs(0, i, d);
  printf("%lld\n", ans);
  return 0;
}

```