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;
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first != b.first)
    return a.first > b.first;
  else
    return a.second > b.second;
}
long long pow1(long long a, long long b) {
  long long ans = a;
  for (long long i = 2; i <= b; i++) ans *= a;
  return ans;
}
int main() {
  long long n;
  cin >> n;
  long long x = n * 5;
  long long low = 0;
  long long high = x;
  long long ans1 = 1e7;
  while (low <= high) {
    long long mid = (low + high) / 2;
    long long t = 0;
    for (long long i = 1; i <= 9; i++) {
      long long temp = pow1(5, i);
      if (temp <= mid) {
        t += (mid / temp);
      } else
        break;
    }
    if (t >= n) {
      ans1 = mid;
      high = mid - 1;
    } else if (t < n)
      low = mid + 1;
  }
  long long t1 = 0;
  for (long long i = 1; i <= 9; i++) {
    long long temp = pow1(5, i);
    if (temp <= ans1) {
      long long d = (ans1 / temp);
      t1 += d;
    } else
      break;
  }
  if (t1 == n) {
    cout << 5 << "\n";
    for (long long i = 0; i < 5; i++) cout << low + i << ' ';
    cout << "\n";
  } else
    cout << 0 << "\n";
  return 0;
}