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 double eps = 1e-15;
const double Max = 1e+18;
inline void normal(long long& a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
inline long long mul(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long add(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long sub(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
int a[555555];
int n, m, ans[55], l;
int main() {
  ios::sync_with_stdio(false);
  for (int i = 5; i < 555555; i *= 5)
    for (int j = i; j < 555555; j += i) a[j]++;
  scanf("%d", &n);
  l = 0;
  for (int i = 1, tot = 0; i < 555555 && tot <= n; i++) {
    tot += a[i];
    if (tot == n) ans[l++] = i;
  }
  printf("%d\n", l);
  for (int i = 0; i < l; i++) printf("%d%c", ans[i], i < l - 1 ? ' ' : '\n');
  return 0;
}