#include <bits/stdc++.h>
using namespace std;
int m, a[100005];
void read();
void solve();
void write();
int main() {
  read();
  solve();
  write();
  return 0;
}
void read() { scanf("%d", &m); }
void solve() {
  int poz = 0;
  for (int i = 1; i <= m; i++) {
    int j = 5 * i;
    int cj = j;
    int nr = 0;
    while (j % 5 == 0) {
      nr++;
      j /= 5;
    }
    poz += nr;
    if (poz <= 100000) {
      a[poz] = cj;
    }
  }
}
void write() {
  if (a[m]) {
    printf("5\n");
    for (int i = a[m]; i <= a[m] + 4; i++) {
      printf("%d ", i);
    }
  } else {
    printf("0");
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 m, a[100005];
void read();
void solve();
void write();
int main() {
  read();
  solve();
  write();
  return 0;
}
void read() { scanf("%d", &m); }
void solve() {
  int poz = 0;
  for (int i = 1; i <= m; i++) {
    int j = 5 * i;
    int cj = j;
    int nr = 0;
    while (j % 5 == 0) {
      nr++;
      j /= 5;
    }
    poz += nr;
    if (poz <= 100000) {
      a[poz] = cj;
    }
  }
}
void write() {
  if (a[m]) {
    printf("5\n");
    for (int i = a[m]; i <= a[m] + 4; i++) {
      printf("%d ", i);
    }
  } else {
    printf("0");
  }
}

```