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;
int n, f, E, temp1, temp2;
long long A[10004];
int main() {
  long long X;
  cin >> X;
  long long c = 0;
  int B = 0;
  int total = 0;
  for (long long i = 1; i <= 1000000000000; i++) {
    long long E = i;
    while ((E % 5) == 0) {
      E /= 5;
      total++;
    }
    if (total == X) {
      A[c] = i;
      c++;
    }
    if (total > X) break;
  }
  cout << c << endl;
  for (long long i = 0; i < c; i++) {
    if (i == 0) {
      cout << A[i];
    } else
      cout << ' ' << A[i];
  }
  cout << endl;
  return 0;
}