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;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroint-loops")
long long int power(long long int b, long long int e, long long int m) {
  if (b == 0) return 0;
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b % m, e / 2, m) % m;
  return power(b * b % m, e / 2, m);
}
long long int power(long long int b, long long int e) {
  if (b == 0) return 0;
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b, e / 2);
  return power(b * b, e / 2);
}
bool isPowerOfTwo(long long int x) { return (x && !(x & (x - 1))); }
long long int m;
bool check(long long int n) {
  long long int cnt = 0, curr = 5;
  while (n >= curr) {
    cnt += n / curr;
    curr *= 5;
  }
  return cnt >= m;
}
long long int get(long long int n) {
  long long int cnt = 0, curr = 5;
  while (n >= curr) {
    cnt += n / curr;
    curr *= 5;
  }
  return cnt;
}
void solve() {
  cin >> m;
  long long int low = 0, high = 1e18, ans = 0;
  while (low <= high) {
    long long int mid = low + (high - low) / 2;
    if (check(mid)) {
      ans = mid;
      high = mid - 1;
    } else {
      low = mid + 1;
    }
  }
  if (get(ans) == m) {
    cout << 5 << "\n";
    for (long long int i = 0; i < 5; i++) cout << ans + i << " ";
  } else {
    cout << 0;
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int tt = 1;
  while (tt--) {
    solve();
    cout << "\n";
  }
  return 0;
}