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 double eps = 1e-6;
const double PI = acos(-1.0);
int num[500000];
int a[11111];
int ok(int n) {
  int ans = 0;
  while (n) {
    n /= 5;
    ans += n;
  }
  return ans;
}
int main() {
  for (int i = 0; i < 500000; i++) {
    num[i] = ok(i);
  }
  int n, Num;
  cin >> n;
  Num = 0;
  for (int i = 0; i < 500000; i++) {
    if (num[i] == n) {
      a[Num++] = i;
    }
  }
  printf("%d\n", Num);
  if (Num) {
    for (int i = 0; i < Num; i++) {
      if (i) printf(" ");
      printf("%d", a[i]);
    }
    printf("\n");
  }
  return 0;
}