#include <bits/stdc++.h>
using namespace std;
int n, l, r, flag, cnt;
int cek(int x) {
  int ret;
  ret = 0;
  while (x > 0) {
    ret += x / 5;
    x /= 5;
  }
  return ret;
}
int main() {
  cin >> n;
  l = 1;
  r = 500000;
  flag = 0;
  while (l < r) {
    int teng, cteng;
    teng = (l + r) / 2;
    cteng = cek(teng);
    if (cteng < n) {
      l = teng + 1;
    } else if (cteng > n) {
      r = teng - 1;
    } else {
      flag = teng;
      while (flag % 5 != 0) {
        flag--;
      }
      break;
    }
  }
  if (l >= r) {
    cout << 0 << endl;
    return 0;
  }
  cout << 5 << endl;
  for (int i = 0; i < 5; i++) {
    cout << flag + i;
    if (i == 4)
      cout << endl;
    else
      cout << " ";
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 n, l, r, flag, cnt;
int cek(int x) {
  int ret;
  ret = 0;
  while (x > 0) {
    ret += x / 5;
    x /= 5;
  }
  return ret;
}
int main() {
  cin >> n;
  l = 1;
  r = 500000;
  flag = 0;
  while (l < r) {
    int teng, cteng;
    teng = (l + r) / 2;
    cteng = cek(teng);
    if (cteng < n) {
      l = teng + 1;
    } else if (cteng > n) {
      r = teng - 1;
    } else {
      flag = teng;
      while (flag % 5 != 0) {
        flag--;
      }
      break;
    }
  }
  if (l >= r) {
    cout << 0 << endl;
    return 0;
  }
  cout << 5 << endl;
  for (int i = 0; i < 5; i++) {
    cout << flag + i;
    if (i == 4)
      cout << endl;
    else
      cout << " ";
  }
  return 0;
}

```