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 zcount(int n) {
  int i = 5, ans = 0;
  while (i <= n) {
    ans += n / i;
    i *= 5;
  }
  return ans;
}
int bin_search(int n, int l, int h) {
  if (l >= h) return 0;
  int mid = (l + h) / 2;
  if (n == zcount(mid))
    return mid;
  else if (n > zcount(mid))
    bin_search(n, mid + 1, h);
  else
    bin_search(n, l, mid);
}
int main() {
  int m;
  cin >> m;
  int x = bin_search(m, 1, 1000000);
  vector<int> v;
  for (long long i = x - 5; i <= x + 5; i++) {
    if (zcount(i) == m) v.push_back(i);
  }
  int sz = v.size();
  if (sz) {
    cout << sz << endl;
    for (auto i : v) cout << i << " ";
  } else
    cout << sz;
  cout << endl;
  return 0;
}