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;
const int N = 1e5 + 7;
const int mod = 1e9 + 7;
long long l, r, mid, L, R;
long long calc(long long x) {
  long long cur = 5, res = 0;
  while (cur <= x) res += x / cur, cur *= 5;
  return res;
}
int main() {
  int m;
  cin >> m;
  l = 0, r = 1e18L, mid;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (calc(mid) >= m)
      r = mid - 1, L = mid;
    else
      l = mid + 1;
  }
  l = 0, r = 1e18L, mid;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (calc(mid) <= m)
      l = mid + 1, R = mid;
    else
      r = mid - 1;
  }
  vector<long long> ans;
  for (long long i = L; i <= R; ++i) ans.push_back(i);
  cout << (int)(ans.size()) << "\n";
  for (long long x : ans) cout << x << " ";
  return 0;
}