#include <bits/stdc++.h>
using namespace std;
int x[2000000], p[100];
int main() {
  long long g, k, sum, r, m, fl;
  for (int i = 4; i < 1000000; i++) {
    g = i;
    r = 5;
    sum = 0;
    while (g / r != 0) {
      sum += g / r;
      r *= 5;
    }
    x[i] = sum;
  }
  while (cin >> m) {
    k = 0;
    for (int i = 0; i < 1000000; i++) {
      if (x[i] == m) {
        p[k] = i;
        k++;
      }
      if (x[i] > m) break;
    }
    if (k > 0) {
      cout << k << endl;
      for (int i = 0; i < k; i++) {
        cout << p[i] << " ";
      }
      cout << endl;
    } else {
      cout << 0 << endl;
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
int x[2000000], p[100];
int main() {
  long long g, k, sum, r, m, fl;
  for (int i = 4; i < 1000000; i++) {
    g = i;
    r = 5;
    sum = 0;
    while (g / r != 0) {
      sum += g / r;
      r *= 5;
    }
    x[i] = sum;
  }
  while (cin >> m) {
    k = 0;
    for (int i = 0; i < 1000000; i++) {
      if (x[i] == m) {
        p[k] = i;
        k++;
      }
      if (x[i] > m) break;
    }
    if (k > 0) {
      cout << k << endl;
      for (int i = 0; i < k; i++) {
        cout << p[i] << " ";
      }
      cout << endl;
    } else {
      cout << 0 << endl;
    }
  }
  return 0;
}

```