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 main() {
  int m, t, sum, sum1 = 0;
  cin >> m;
  for (int i = 5; i <= 500000; i = i + 5) {
    t = i;
    sum = 0;
    while (t >= 5) {
      t = t / 5;
      sum++;
      if (t % 5 != 0) {
        break;
      }
    }
    sum = sum + sum1;
    if (sum == m) {
      cout << "5" << endl;
      for (int j = 0; j < 5; j++) {
        cout << i + j << " ";
      }
      break;
    } else if (sum > m) {
      cout << "0" << endl;
      break;
    }
    sum1 = sum;
  }
}