#include <bits/stdc++.h>
using namespace std;
const int p[] = {5, 25, 125, 625, 3125, 15625, 78125, 390625, 1953125};
int c0(int x) {
  int ans = 0;
  for (int i = 0; i < 9; i++) ans += (x / p[i]);
  return ans;
}
int main() {
  int m, l = 1, r = 500000;
  scanf("%d", &m);
  while (r - l > 1) {
    int x = (r + l) / 2;
    if (c0(x) < m)
      l = x;
    else
      r = x;
  }
  if (c0(r) == m)
    printf("5\n%d %d %d %d %d", r, r + 1, r + 2, r + 3, r + 4);
  else
    printf("0");
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial of n ends with exactly m zeroes. Are you among those great programmers who can solve this problem?

Input

The only line of input contains an integer m (1 ≤ m ≤ 100 000) — the required number of trailing zeroes in factorial.

Output

First print k — the number of values of n such that the factorial of n ends with m zeroes. Then print these k integers in increasing order.

Examples

Input

1


Output

5
5 6 7 8 9 

Input

5


Output

0

Note

The factorial of n is equal to the product of all integers from 1 to n inclusive, that is n! = 1·2·3·...·n.

In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int p[] = {5, 25, 125, 625, 3125, 15625, 78125, 390625, 1953125};
int c0(int x) {
  int ans = 0;
  for (int i = 0; i < 9; i++) ans += (x / p[i]);
  return ans;
}
int main() {
  int m, l = 1, r = 500000;
  scanf("%d", &m);
  while (r - l > 1) {
    int x = (r + l) / 2;
    if (c0(x) < m)
      l = x;
    else
      r = x;
  }
  if (c0(r) == m)
    printf("5\n%d %d %d %d %d", r, r + 1, r + 2, r + 3, r + 4);
  else
    printf("0");
  return 0;
}

```