#include <bits/stdc++.h>
using namespace std;
const int INF = 2e9;
const long long MOD = 1e9 + 7;
const int MAXN = 1e5 + 2;
int ok(int n) {
  int res = 0;
  while (n) {
    res += n / 5;
    n /= 5;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int m;
  cin >> m;
  int lo = 1, hi = 1e6;
  int mid1 = -1;
  while (lo < hi) {
    mid1 = (lo + hi) >> 1;
    if (ok(mid1) >= m)
      hi = mid1;
    else
      lo = mid1 + 1;
  }
  mid1 = lo;
  int mid2 = -1;
  for (int i = mid1; i < mid1 + 10; ++i)
    if (ok(i) >= (m + 1))
      break;
    else
      mid2 = i;
  if (ok(mid1) != m) mid2 = -1;
  if (mid2 == -1)
    cout << 0 << endl;
  else {
    cout << mid2 - mid1 + 1 << endl;
    for (int i = mid1; i <= mid2; ++i) cout << i << " ";
    cout << endl;
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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 int INF = 2e9;
const long long MOD = 1e9 + 7;
const int MAXN = 1e5 + 2;
int ok(int n) {
  int res = 0;
  while (n) {
    res += n / 5;
    n /= 5;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int m;
  cin >> m;
  int lo = 1, hi = 1e6;
  int mid1 = -1;
  while (lo < hi) {
    mid1 = (lo + hi) >> 1;
    if (ok(mid1) >= m)
      hi = mid1;
    else
      lo = mid1 + 1;
  }
  mid1 = lo;
  int mid2 = -1;
  for (int i = mid1; i < mid1 + 10; ++i)
    if (ok(i) >= (m + 1))
      break;
    else
      mid2 = i;
  if (ok(mid1) != m) mid2 = -1;
  if (mid2 == -1)
    cout << 0 << endl;
  else {
    cout << mid2 - mid1 + 1 << endl;
    for (int i = mid1; i <= mid2; ++i) cout << i << " ";
    cout << endl;
  }
  return 0;
}

```