#include <bits/stdc++.h>
using namespace std;
int zeroes[1 << 20];
int compute(int n) {
  int pw = 5;
  int z = 0;
  while (1) {
    z += n / pw;
    pw *= 5;
    if (pw > n) return z;
  }
}
int main() {
  int m;
  scanf("%d", &m);
  for (int i = 0; i <= 1000000; i++) zeroes[i] = compute(i);
  int cnt = 0;
  for (int i = 0; i <= 1000000; i++)
    if (zeroes[i] == m) cnt++;
  printf("%d\n", cnt);
  for (int i = 0; i <= 1000000; i++)
    if (zeroes[i] == m) printf("%d ", i);
  puts("");
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
int zeroes[1 << 20];
int compute(int n) {
  int pw = 5;
  int z = 0;
  while (1) {
    z += n / pw;
    pw *= 5;
    if (pw > n) return z;
  }
}
int main() {
  int m;
  scanf("%d", &m);
  for (int i = 0; i <= 1000000; i++) zeroes[i] = compute(i);
  int cnt = 0;
  for (int i = 0; i <= 1000000; i++)
    if (zeroes[i] == m) cnt++;
  printf("%d\n", cnt);
  for (int i = 0; i <= 1000000; i++)
    if (zeroes[i] == m) printf("%d ", i);
  puts("");
  return 0;
}

```