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;
const long long int N = 5 * 1e5 + 5;
const int mod = 1073741824;
int h[N] = {0};
long long expo(long long x, long long n, long long mod) {
  long long res = 1LL;
  while (n) {
    if (n & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    n = n / 2;
  }
  return res;
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int i, c = 0;
  int h[N] = {0};
  for (int l = 5; l < N; l += 5) {
    c = 0;
    i = l;
    while (i % 5 == 0) i /= 5, c++;
    h[l] = c;
  }
  for (int i = (1); i < N; i++) h[i] += h[i - 1];
  int n;
  cin >> n;
  int x = lower_bound(h, h + N, n) - h;
  if (h[x] == n) {
    cout << 5 << endl;
    for (int i = (0); i < 5; i++) cout << x + i << ' ';
  } else
    cout << 0;
  return 0;
}