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;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const long long INF = 1e16;
long long m;
long long get(long long x) {
  long long ret = 0;
  while (x > 0) {
    ret += x / 5;
    x /= 5;
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> m;
  long long l = 5, r = INF, ans = -1;
  while (l <= r) {
    long long mid = (l + r) >> 1;
    if (get(mid) >= m) {
      r = mid - 1;
      ans = mid;
    } else {
      l = mid + 1;
    }
  }
  if (get(ans) == m) {
    cout << 5 << '\n';
    for (int i = 0; i < 5; i++) {
      cout << ans + i << (i == 4 ? '\n' : ' ');
    }
  } else {
    cout << 0 << '\n';
  }
  return 0;
}