#include <bits/stdc++.h>
const int INF = 2147483647;
const int MOD = 1000000007;
const long long INF64 = 9223372036854775807;
const long double EPS = 1e-7;
using namespace std;
int get5(int n) {
  int ans = 0;
  while (n % 5 == 0) {
    ans++;
    n /= 5;
  }
  return ans;
}
int get2(int n) {
  int ans = 0;
  while (n % 2 == 0) {
    ans++;
    n /= 2;
  }
  return ans;
}
int main() {
  int n, i, cnt = 0, k, cnt2 = 0, cnt5 = 0;
  vector<int> ans;
  scanf("%d", &n);
  for (i = 1; cnt <= n; i++) {
    cnt2 += get2(i);
    cnt5 += get5(i);
    k = min(cnt2, cnt5);
    cnt += k;
    cnt2 -= k;
    cnt5 -= k;
    if (cnt == n) ans.push_back(i);
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
const int INF = 2147483647;
const int MOD = 1000000007;
const long long INF64 = 9223372036854775807;
const long double EPS = 1e-7;
using namespace std;
int get5(int n) {
  int ans = 0;
  while (n % 5 == 0) {
    ans++;
    n /= 5;
  }
  return ans;
}
int get2(int n) {
  int ans = 0;
  while (n % 2 == 0) {
    ans++;
    n /= 2;
  }
  return ans;
}
int main() {
  int n, i, cnt = 0, k, cnt2 = 0, cnt5 = 0;
  vector<int> ans;
  scanf("%d", &n);
  for (i = 1; cnt <= n; i++) {
    cnt2 += get2(i);
    cnt5 += get5(i);
    k = min(cnt2, cnt5);
    cnt += k;
    cnt2 -= k;
    cnt5 -= k;
    if (cnt == n) ans.push_back(i);
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}

```