#include <bits/stdc++.h>
using namespace std;
string tostr(int n) {
  stringstream rr;
  rr << n;
  return rr.str();
}
long long power(long long a, long long b) {
  return b == 0 ?: a * power(a, b - 1);
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long solve(long long n, long long b) {
  long long ans = 1000000000000000000LL;
  for (long long i = 2; i * i <= b; i++) {
    int cnt = 0;
    while (b % i == 0) {
      b = b / i;
      cnt++;
    }
    if (cnt == 0) continue;
    long long tmp = 1, sum = 0;
    while (tmp <= n / i) {
      tmp *= i;
      sum += (n / tmp);
    }
    ans = min(ans, sum / cnt);
  }
  if (b != 1) {
    long long tmp = 1, sum = 0;
    while (tmp <= n / b) {
      tmp *= b;
      sum += (n / tmp);
    }
    ans = min(ans, sum);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  long long m;
  while (cin >> m) {
    vector<long long> zero;
    zero.push_back(0);
    for (int i = 1; i <= 500000; i++) zero.push_back(solve(i, 10));
    if (find(zero.begin(), zero.end(), m) == zero.end())
      cout << 0 << endl;
    else {
      long long ans;
      cout << 5 << endl;
      for (int i = 0; i < zero.size(); i++) {
        if (zero[i] == m) {
          ans = i;
          break;
        }
      }
      cout << ans << " " << ans + 1 << " " << ans + 2 << " " << ans + 3 << " "
           << ans + 4 << endl;
    }
  }
  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;
string tostr(int n) {
  stringstream rr;
  rr << n;
  return rr.str();
}
long long power(long long a, long long b) {
  return b == 0 ?: a * power(a, b - 1);
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long solve(long long n, long long b) {
  long long ans = 1000000000000000000LL;
  for (long long i = 2; i * i <= b; i++) {
    int cnt = 0;
    while (b % i == 0) {
      b = b / i;
      cnt++;
    }
    if (cnt == 0) continue;
    long long tmp = 1, sum = 0;
    while (tmp <= n / i) {
      tmp *= i;
      sum += (n / tmp);
    }
    ans = min(ans, sum / cnt);
  }
  if (b != 1) {
    long long tmp = 1, sum = 0;
    while (tmp <= n / b) {
      tmp *= b;
      sum += (n / tmp);
    }
    ans = min(ans, sum);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  long long m;
  while (cin >> m) {
    vector<long long> zero;
    zero.push_back(0);
    for (int i = 1; i <= 500000; i++) zero.push_back(solve(i, 10));
    if (find(zero.begin(), zero.end(), m) == zero.end())
      cout << 0 << endl;
    else {
      long long ans;
      cout << 5 << endl;
      for (int i = 0; i < zero.size(); i++) {
        if (zero[i] == m) {
          ans = i;
          break;
        }
      }
      cout << ans << " " << ans + 1 << " " << ans + 2 << " " << ans + 3 << " "
           << ans + 4 << endl;
    }
  }
  return 0;
}

```