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 m;
const int maxn = 100100;
long long int arr[maxn];
int k = 0;
int main() {
  cin >> m;
  for (long long int i = 0; i < 1000000; i++) {
    long long int sk = i;
    long long int cur = 5;
    int cnt = 0;
    while (1) {
      sk /= cur;
      if (sk == 0) {
        break;
      }
      cnt += sk;
      sk = i;
      cur *= 5;
    }
    if (cnt == m) {
      arr[k++] = i;
    }
  }
  cout << k << endl;
  for (int i = 0; i < k; i++) {
    cout << arr[i] << " ";
  }
  cout << endl;
  return 0;
}