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;
int ceros(int num) {
  int i, r = 0;
  for (i = 1; pow(5, i) <= num; i++) {
    r = r + floor(num / pow(5, i));
  }
  return r;
}
int main() {
  long int n, m, i;
  vector<int> cumplen;
  cin >> m;
  for (i = 1; i < 1000000; i++) {
    if (ceros(i) == m) cumplen.push_back(i);
  }
  cout << cumplen.size() << endl;
  while (cumplen.size()) {
    cout << cumplen[0];
    cumplen.erase(cumplen.begin());
    if (!cumplen.empty()) cout << " ";
  }
  return 0;
}