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 int N = 1000001;
int dp[N] = {0, 5};
int f(int x) {
  int ans = 0;
  while (x) {
    ans += x / 5;
    x /= 5;
  }
  return ans;
}
int main(void) {
  int m;
  int tmp = 1;
  int i;
  for (i = 5; i <= 400005; i += 5) {
    int tmp = f(i);
    dp[tmp] = i;
  }
  while (cin >> m) {
    if (!dp[m]) {
      cout << 0 << endl;
      continue;
    }
    cout << 5 << endl;
    for (i = dp[m]; i < dp[m] + 5; i++) cout << i << ' ';
    cout << endl;
  }
  return 0;
}