#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1e9 + 7;
struct custom {
  bool operator()(const pair<int, int> &p1, const pair<int, int> &p2) const {
    return p1.first < p2.first;
  }
};
long long int get_pow(long long int x, long long int k) {
  if (k == 0) return 1ll;
  long long int y = get_pow(x, k / 2);
  y = (y * y) % 60;
  if (k % 2) y = (y * x) % 60;
  return y;
}
void multiply(long long int f[][2], long long int g[][2]) {
  long long int a = (f[0][0] * g[0][0] + f[0][1] * g[1][0]) % 10;
  long long int b = (f[0][0] * g[0][1] + f[0][1] * g[1][1]) % 10;
  long long int c = (f[1][0] * g[0][0] + f[1][1] * g[1][0]) % 10;
  long long int d = (f[1][0] * g[0][1] + f[1][1] * g[1][1]) % 10;
  f[0][0] = a;
  f[0][1] = b;
  f[1][0] = c;
  f[1][1] = d;
}
void power(long long int f[2][2], long long int n) {
  long long int g[2][2] = {{1, 1}, {1, 0}};
  if (n == 0 || n == 1) return;
  power(f, n / 2);
  multiply(f, f);
  if (n % 2 == 1) multiply(f, g);
}
long long int fib(long long int n) {
  long long int f[2][2] = {{1, 1}, {1, 0}};
  if (n == 0) return 0;
  power(f, n - 1);
  return f[0][0] % 10;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int test = 1;
  for (int tes = (int)0; tes < int(test); tes++) {
    long long int n, done = 0, m = 0;
    cin >> n;
    long long int val = 5, z = 5;
    vector<long long> v;
    for (int i = (int)0; i < int(30); i++) {
      v.push_back(z);
      z = z * 5;
    }
    while (true) {
      long long int d = 0;
      for (int i = (int)0; i < int(v.size()); i++) {
        if (val % v[i] == 0) d = i + 1;
      }
      m += d;
      if (m == n) {
        done = 1;
        break;
      } else if (m > n)
        break;
      val += 5;
    }
    if (!done) {
      cout << 0 << "\n";
      continue;
    }
    cout << 5 << "\n";
    for (int i = (int)0; i < int(5); i++) cout << val + i << " ";
    cout << "\n";
  }
}

### 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>
using namespace std;
const long long int MOD = 1e9 + 7;
struct custom {
  bool operator()(const pair<int, int> &p1, const pair<int, int> &p2) const {
    return p1.first < p2.first;
  }
};
long long int get_pow(long long int x, long long int k) {
  if (k == 0) return 1ll;
  long long int y = get_pow(x, k / 2);
  y = (y * y) % 60;
  if (k % 2) y = (y * x) % 60;
  return y;
}
void multiply(long long int f[][2], long long int g[][2]) {
  long long int a = (f[0][0] * g[0][0] + f[0][1] * g[1][0]) % 10;
  long long int b = (f[0][0] * g[0][1] + f[0][1] * g[1][1]) % 10;
  long long int c = (f[1][0] * g[0][0] + f[1][1] * g[1][0]) % 10;
  long long int d = (f[1][0] * g[0][1] + f[1][1] * g[1][1]) % 10;
  f[0][0] = a;
  f[0][1] = b;
  f[1][0] = c;
  f[1][1] = d;
}
void power(long long int f[2][2], long long int n) {
  long long int g[2][2] = {{1, 1}, {1, 0}};
  if (n == 0 || n == 1) return;
  power(f, n / 2);
  multiply(f, f);
  if (n % 2 == 1) multiply(f, g);
}
long long int fib(long long int n) {
  long long int f[2][2] = {{1, 1}, {1, 0}};
  if (n == 0) return 0;
  power(f, n - 1);
  return f[0][0] % 10;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int test = 1;
  for (int tes = (int)0; tes < int(test); tes++) {
    long long int n, done = 0, m = 0;
    cin >> n;
    long long int val = 5, z = 5;
    vector<long long> v;
    for (int i = (int)0; i < int(30); i++) {
      v.push_back(z);
      z = z * 5;
    }
    while (true) {
      long long int d = 0;
      for (int i = (int)0; i < int(v.size()); i++) {
        if (val % v[i] == 0) d = i + 1;
      }
      m += d;
      if (m == n) {
        done = 1;
        break;
      } else if (m > n)
        break;
      val += 5;
    }
    if (!done) {
      cout << 0 << "\n";
      continue;
    }
    cout << 5 << "\n";
    for (int i = (int)0; i < int(5); i++) cout << val + i << " ";
    cout << "\n";
  }
}

```