#include <bits/stdc++.h>
using namespace std;
long long int n = 1000001;
vector<long long int> v(1000001);
void sieve() {
  for (int i = 5; i < n; ++i) {
    int two_count = 0, five_count = 0;
    int temp, j = 1;
    do {
      temp = i / (int)pow(2, j);
      two_count += temp;
      j++;
    } while (temp > 0);
    j = 1;
    do {
      temp = i / (int)pow(5, j);
      five_count += temp;
      j++;
    } while (temp > 0);
    v[i] = min(two_count, five_count);
  }
}
int main() {
  sieve();
  int k;
  vector<int> ans;
  cin >> k;
  for (int i = 5; i < n; i++) {
    if (v[i] == k) ans.push_back(i);
  }
  if (ans.size() == 0)
    cout << 0;
  else {
    cout << ans.size() << endl;
    for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  }
  return (0);
}

### Prompt

Create a solution in CPP for the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int n = 1000001;
vector<long long int> v(1000001);
void sieve() {
  for (int i = 5; i < n; ++i) {
    int two_count = 0, five_count = 0;
    int temp, j = 1;
    do {
      temp = i / (int)pow(2, j);
      two_count += temp;
      j++;
    } while (temp > 0);
    j = 1;
    do {
      temp = i / (int)pow(5, j);
      five_count += temp;
      j++;
    } while (temp > 0);
    v[i] = min(two_count, five_count);
  }
}
int main() {
  sieve();
  int k;
  vector<int> ans;
  cin >> k;
  for (int i = 5; i < n; i++) {
    if (v[i] == k) ans.push_back(i);
  }
  if (ans.size() == 0)
    cout << 0;
  else {
    cout << ans.size() << endl;
    for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  }
  return (0);
}

```