#include <bits/stdc++.h>
#pragma comment(linker, "/stack:16777216")
using namespace std;
const int INF = 1000000000;
const int BASE = 1000000007;
const int MAX = 1000007;
const int MAX2 = 7777;
const int MAXE = 100000;
const int ADD = 1000000;
const int MOD = 1000000007;
const int CNT = 800;
int dp[MAX];
int main() {
  int m;
  cin >> m;
  vector<int> res;
  for (int i = (1); i < (MAX); i++) {
    dp[i] = dp[i / 5] + i / 5;
    if (dp[i] == m) res.push_back(i);
  }
  cout << res.size() << endl;
  for (int i = (0); i < (res.size()); i++) {
    printf("%d ", res[i]);
  }
  return 0;
}

### Prompt

Please formulate a cpp solution to 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>
#pragma comment(linker, "/stack:16777216")
using namespace std;
const int INF = 1000000000;
const int BASE = 1000000007;
const int MAX = 1000007;
const int MAX2 = 7777;
const int MAXE = 100000;
const int ADD = 1000000;
const int MOD = 1000000007;
const int CNT = 800;
int dp[MAX];
int main() {
  int m;
  cin >> m;
  vector<int> res;
  for (int i = (1); i < (MAX); i++) {
    dp[i] = dp[i / 5] + i / 5;
    if (dp[i] == m) res.push_back(i);
  }
  cout << res.size() << endl;
  for (int i = (0); i < (res.size()); i++) {
    printf("%d ", res[i]);
  }
  return 0;
}

```