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 v[1000009];
int main() {
  v[0] = v[1] = v[2] = v[3] = v[4] = 0;
  for (int i = 5; i < 1000009; i += 5) {
    int ct = 0, x = i;
    while (x % 5 == 0) {
      x /= 5;
      ct++;
    }
    v[i] = ct + v[i - 5];
    v[i + 1] = v[i + 2] = v[i + 3] = v[i + 4] = v[i];
  }
  int m;
  cin >> m;
  int f = 0;
  vector<int> vc;
  for (int i = 0; i < 1000009; i++) {
    if (v[i] == m) {
      f++;
      vc.push_back(i);
    } else if (v[i] > m)
      break;
  }
  if (f == 0)
    cout << 0;
  else {
    cout << f << endl;
    for (vector<int>::iterator it = vc.begin(); it < vc.end(); it++)
      cout << *it << " ";
  }
  return 0;
}