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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int n;
int a[2000007];
long long int tmp, re, sum, dem1, dem2;
void _love() { cin >> n; }
void _nang_am() {
  for (long long int i = 1; i <= 1e6; i++) {
    tmp = i;
    while (tmp % 2 == 0) {
      dem1++;
      tmp /= 2;
    }
    while (tmp % 5 == 0) {
      dem2++;
      tmp /= 5;
    }
    dem2 = min(dem2, dem1);
    if (dem2 == n) {
      cout << 5 << "\n";
      for (long long int j = 1; j <= 5; j++) {
        cout << i + j - 1 << " ";
      }
      return;
    }
  }
  cout << "0"
       << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  _love();
  _nang_am();
  return 0;
}