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;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const int msz = 10000;
int main() {
  int m;
  scanf("%d", &m);
  int n = 0;
  int st = 5;
  m--;
  bool f = 0;
  while (m) {
    st += 5;
    int j = st;
    while (m && j % 5 == 0) {
      j /= 5;
      --m;
    }
    if (j % 5 == 0) {
      f = 1;
      break;
    }
  }
  if (f) {
    puts("0");
    return 0;
  }
  printf("%d\n", 5);
  for (int i = st; i < st + 5; ++i) printf("%d ", i);
  return 0;
}