A Trivial 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:

#include <bits/stdc++.h>

using namespace std;

inline int get(int n) {
int z = 0;
while (n > 0) {
n /= 5;
z += n;
}
return z;
}

int main() {
int k;
scanf("%d", &k);
int low = 0, high = (int) 1e9;
while (low < high) {
    int mid = (low + high) >> 1;
int cnt = get(mid);
if (cnt < k) {
      low = mid + 1;
    } else {
      high = mid;
    }
  }
  vector <int> res;
while (get(low) == k) {
res.push_back(low);
low++;
}
int sz = res.size();
printf("%d\n", sz);
for (int i = 0; i < sz; i++) {
    if (i > 0) {
putchar(' ');
}
printf("%d", res[i]);
}
printf("\n");
return 0;
}