#include <bits/stdc++.h>
long long MOD = 1000000007;
using namespace std;
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
long long POW(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a % MOD;
  long long temp = POW(a, b / 2);
  if (b % 2 == 0)
    return (temp * temp) % MOD;
  else
    return (((temp * temp) % MOD) * a) % MOD;
}
vector<char> capl = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I',
                     'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R',
                     'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};
vector<char> sml = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i',
                    'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r',
                    's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  map<int, int> p;
  int d = 0;
  for (int i = 5; i < 1000000; i = i + 5) {
    int c = i;
    while (c % 5 == 0) {
      c = c / 5;
      d++;
    }
    p[d] = i;
  }
  int m;
  cin >> m;
  if (p[m] == 0) {
    cout << "0";
  } else {
    cout << "5\n";
    cout << p[m] << " " << ++p[m] << " " << ++p[m] << " " << ++p[m] << " "
         << ++p[m];
  }
}

### Prompt

Develop a solution in Cpp 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>
long long MOD = 1000000007;
using namespace std;
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
long long POW(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a % MOD;
  long long temp = POW(a, b / 2);
  if (b % 2 == 0)
    return (temp * temp) % MOD;
  else
    return (((temp * temp) % MOD) * a) % MOD;
}
vector<char> capl = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I',
                     'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R',
                     'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};
vector<char> sml = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i',
                    'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r',
                    's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  map<int, int> p;
  int d = 0;
  for (int i = 5; i < 1000000; i = i + 5) {
    int c = i;
    while (c % 5 == 0) {
      c = c / 5;
      d++;
    }
    p[d] = i;
  }
  int m;
  cin >> m;
  if (p[m] == 0) {
    cout << "0";
  } else {
    cout << "5\n";
    cout << p[m] << " " << ++p[m] << " " << ++p[m] << " " << ++p[m] << " "
         << ++p[m];
  }
}

```