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;
long long power(long long, long long);
long long mpower(long long, long long);
vector<long long> v;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (long long i = 1; i <= 1000000; i++) {
    long long cnt = 0;
    long long t = 5;
    while (i >= t) {
      cnt += i / t;
      t = t * 5;
    }
    v.push_back(cnt);
  }
  long long n;
  cin >> n;
  vector<long long> ans;
  for (long long i = 1; i <= 1000000; i++) {
    if (v[i - 1] == n) ans.push_back(i);
    if (v[i - 1] > n) break;
  }
  cout << ans.size() << endl;
  for (long long i : ans) cout << i << " ";
  return 0;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 == 1) res = res * a;
    a = a * a;
    b = b / 2;
  }
  return res;
}
long long mpower(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      res = (res * a) % 1000000007;
    }
    a = (a * a) % 1000000007;
    b = b / 2;
  }
  return res;
}