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;
int prime[1 << 20];
int numofprimes;
int gcd(int a, int b) { return (b == 0 ? a : gcd(b, a % b)); }
int lcm(int a, int b) { return (a * (b / gcd(a, b))); }
int POWER(int x, int y, int m) {
  if (y == 0) return 1;
  if (y % 2 == 0) {
    int temp = POWER(x, y / 2, m);
    temp *= temp;
    temp %= m;
    return temp;
  }
  int temp = POWER(x, y - 1, m);
  temp *= x;
  temp %= m;
  return temp;
}
void runEratosthenesSieve(int upperBound) {
  int upperBoundSquareRoot = (int)sqrt((double)upperBound);
  bool *isComposite = new bool[upperBound + 1];
  memset(isComposite, 0, sizeof(bool) * (upperBound + 1));
  for (int m = 2; m <= upperBoundSquareRoot; m++) {
    if (!isComposite[m]) {
      prime[numofprimes++] = m;
      for (int k = m * m; k <= upperBound; k += m) isComposite[k] = true;
    }
  }
  for (int m = upperBoundSquareRoot + 1; m <= upperBound; m++)
    if (!isComposite[m]) prime[numofprimes++];
}
int main() {
  int n;
  cin >> n;
  long long int st = 1;
  for (long long int i = 1;; i += 5) {
    long long int s = 5;
    long long int cnt = 0;
    while (s <= i) {
      cnt += i / s;
      s *= 5;
    }
    if (cnt > n) {
      cout << 0 << '\n';
      return 0;
    }
    if (cnt == n) {
      cout << 5 << '\n';
      cout << i - 1 << " " << i << " " << i + 1 << " " << i + 2 << " " << i + 3
           << '\n';
      return 0;
    }
  }
}