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;
using namespace std;
struct point {
  long long x;
  long long y;
  point(long long a = 0, long long b = 0) {
    x = a;
    y = b;
  }
};
double sqr(double a) { return a * a; }
bool cmp(long long a, long long b) { return a < b; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int m;
  int x, y, a, b;
  x = y = 0;
  cin >> m;
  for (int i = 2; i <= 500000; i++) {
    a = i;
    while (a % 2 == 0) {
      a = a / 2;
      x++;
    }
    while (a % 5 == 0) {
      a = a / 5;
      y++;
    }
    if (min(x, y) == m) {
      a = i;
      break;
    }
    if (min(x, y) > m) {
      cout << 0;
      return 0;
    }
  }
  vector<int> z;
  z.push_back(a);
  b = a;
  for (int i = 0; i < 20; i++) {
    b++;
    a = b;
    while (a % 2 == 0) {
      a = a / 2;
      x++;
    }
    while (a % 5 == 0) {
      a = a / 5;
      y++;
    }
    if (min(x, y) == m) z.push_back(b);
    if (min(x, y) > m) {
      break;
    }
  }
  cout << z.size() << endl;
  for (int i = 0; i < z.size(); i++) {
    cout << z[i] << " ";
  }
  return 0;
}