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;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int abs1(long long int a) {
  if (a < 0)
    return -1 * a;
  else
    return a;
}
long long int power(long long int base, long long int n, long long int M) {
  if (n == 0) return 1;
  if (n == 1) return base;
  long long int halfn = power(base, n / 2, M);
  if (n % 2 == 0)
    return (halfn * halfn) % M;
  else
    return (((halfn * halfn) % M) * base) % M;
}
long long int findMMI(long long int n, long long int M) {
  return power(n, M - 2, M);
}
long long int max2(long long int a, long long int b) {
  if (a > b)
    return a;
  else
    return b;
}
long long int max3(long long int a, long long int b, long long int c) {
  max2(max2(a, b), c);
}
long long int min2(long long int a, long long int b) {
  if (a < b)
    return a;
  else
    return b;
}
long long int min3(long long int a, long long int b, long long int c) {
  return min2(min2(a, b), c);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int m;
  cin >> m;
  long long int ct = 0, fin = 0;
  vector<long long int> v;
  for (int i = 5; i <= 100000000; i++) {
    long long int t = 5, tct = 0;
    while (t <= i) {
      if (i % t == 0) tct++;
      t *= 5;
    }
    ct += tct;
    if (ct == m) v.push_back(i);
    if (ct > m) break;
  }
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++) cout << v[i] << " ";
  return 0;
}