#include <bits/stdc++.h>
using namespace std;
long long int myfind(long long int x) {
  long long int ans = 0, t, d = 5;
  while (1) {
    t = x / d;
    if (t) {
      ans += t;
      d *= 5;
    } else
      break;
  }
  return ans;
}
long long int myfindl(long long int l, long long int h, long long int v) {
  long long int ans = -1, m, x;
  while (l <= h) {
    m = ((h - l) >> 1) + l;
    x = myfind(m);
    if (x == v) {
      ans = m;
      h = m - 1;
    } else if (x < v) {
      l = m + 1;
    } else
      h = m - 1;
  }
  return ans;
}
long long int myfindh(long long int l, long long int h, long long int v) {
  long long int ans = -1, m, x;
  while (l <= h) {
    m = ((h - l) >> 1) + l;
    x = myfind(m);
    if (x == v) {
      ans = m;
      l = m + 1;
    } else if (x < v) {
      l = m + 1;
    } else
      h = m - 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int m;
  cin >> m;
  long long int n, i, ansl, ansh;
  ansl = myfindl(0, LONG_MAX, m);
  if (ansl != -1) {
    ansh = myfindh(0, LONG_MAX, m);
    n = ansh - ansl + 1;
    cout << n << endl;
    for (i = 0; i < n; i++) {
      cout << (ansl + i) << " ";
    }
    cout << endl;
  } else
    cout << 0 << 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;
long long int myfind(long long int x) {
  long long int ans = 0, t, d = 5;
  while (1) {
    t = x / d;
    if (t) {
      ans += t;
      d *= 5;
    } else
      break;
  }
  return ans;
}
long long int myfindl(long long int l, long long int h, long long int v) {
  long long int ans = -1, m, x;
  while (l <= h) {
    m = ((h - l) >> 1) + l;
    x = myfind(m);
    if (x == v) {
      ans = m;
      h = m - 1;
    } else if (x < v) {
      l = m + 1;
    } else
      h = m - 1;
  }
  return ans;
}
long long int myfindh(long long int l, long long int h, long long int v) {
  long long int ans = -1, m, x;
  while (l <= h) {
    m = ((h - l) >> 1) + l;
    x = myfind(m);
    if (x == v) {
      ans = m;
      l = m + 1;
    } else if (x < v) {
      l = m + 1;
    } else
      h = m - 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int m;
  cin >> m;
  long long int n, i, ansl, ansh;
  ansl = myfindl(0, LONG_MAX, m);
  if (ansl != -1) {
    ansh = myfindh(0, LONG_MAX, m);
    n = ansh - ansl + 1;
    cout << n << endl;
    for (i = 0; i < n; i++) {
      cout << (ansl + i) << " ";
    }
    cout << endl;
  } else
    cout << 0 << endl;
  return 0;
}

```