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;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-ffloat-store")
#pragma GCC optimize("-fno-defer-pop")
long long int power(long long int a, long long int b, long long int m) {
  if (b == 0) return 1;
  if (b == 1) return a % m;
  long long int t = power(a, b / 2, m) % m;
  t = (t * t) % m;
  if (b & 1) t = ((t % m) * (a % m)) % m;
  return t;
}
long long int modInverse(long long int a, long long int m) {
  return power(a, m - 2, m);
}
long long int solve(long long int n) {
  long long int a, b, mid, an = -1, k, l;
  a = 1;
  b = 1e9;
  while (b >= a) {
    mid = (a + b) / 2;
    k = mid;
    l = 0;
    while (k > 0) {
      l += (k / 5);
      k = k / 5;
    }
    if (l == n) {
      an = mid;
      b = mid - 1;
    } else if (l > n) {
      b = mid - 1;
    } else {
      a = mid + 1;
    }
  }
  return an;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int i, j, k, l, n;
  cin >> n;
  k = solve(n);
  if (k == -1) {
    cout << "0"
         << "\n";
  } else {
    cout << "5\n";
    for (i = k; i < k + 5; i++) {
      cout << i << " ";
    }
  }
  return 0;
}