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 dp1[100001], dp2[100001], dp3[100001];
int main() {
  ios::sync_with_stdio(false);
  for (long long int z = 0; z < 10000; z++)
    ;
  long long int m, c = 0;
  cin >> m;
  for (long long int z = 0; z < 10000; z++)
    ;
  for (long long int z = 0; z < 10000; z++)
    ;
  for (long long int i = 5; c <= m; i += 5) {
    long long int x = i;
    while (x > 0 && x % 5 == 0) c++, x /= 5;
    if (c == m) {
      cout << "5" << endl;
      for (long long int j = i; j < i + 5; j++) cout << j << " ";
      return 0;
    }
  }
  for (long long int z = 0; z < 10000; z++)
    ;
  for (long long int z = 0; z < 10000; z++)
    ;
  for (long long int z = 0; z < 10000; z++)
    ;
  for (long long int z = 0; z < 10000; z++)
    ;
  for (long long int z = 0; z < 10000; z++)
    ;
  cout << "0";
  return 0;
}