#include <bits/stdc++.h>
using namespace std;
int m;
bool ok;
int get(int x) {
  int lo = 1, hi = 1e9, ans = 0;
  while (lo <= hi) {
    int mid = lo + (hi - lo) / 2;
    int zeroes = 0;
    long long k = 5;
    long long cnt = 0;
    while (k <= mid) cnt += mid / k, k *= 5LL;
    if (cnt == m) ok = true;
    if (cnt >= x) {
      hi = mid - 1;
      if (cnt == x) ans = mid;
    } else
      lo = mid + 1;
  }
  return ans;
}
int get2(int x) {
  int lo = 1, hi = 1e9, ans = 0;
  while (lo <= hi) {
    int mid = lo + (hi - lo) / 2;
    int zeroes = 0;
    long long k = 5;
    long long cnt = 0;
    if (cnt == m) ok = true;
    while (k <= mid) cnt += mid / k, k *= 5LL;
    if (cnt > x)
      hi = mid - 1;
    else
      lo = mid + 1, ans = mid;
  }
  return ans;
}
int main() {
  cin >> m;
  int l = get(m), r = get2(m);
  if (!ok) {
    cout << 0 << endl;
    return 0;
  }
  cout << r - l + 1 << endl;
  for (int i = l; i <= r; i++) cout << i << ' ';
  cout << endl;
  return 0;
}

### 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;
int m;
bool ok;
int get(int x) {
  int lo = 1, hi = 1e9, ans = 0;
  while (lo <= hi) {
    int mid = lo + (hi - lo) / 2;
    int zeroes = 0;
    long long k = 5;
    long long cnt = 0;
    while (k <= mid) cnt += mid / k, k *= 5LL;
    if (cnt == m) ok = true;
    if (cnt >= x) {
      hi = mid - 1;
      if (cnt == x) ans = mid;
    } else
      lo = mid + 1;
  }
  return ans;
}
int get2(int x) {
  int lo = 1, hi = 1e9, ans = 0;
  while (lo <= hi) {
    int mid = lo + (hi - lo) / 2;
    int zeroes = 0;
    long long k = 5;
    long long cnt = 0;
    if (cnt == m) ok = true;
    while (k <= mid) cnt += mid / k, k *= 5LL;
    if (cnt > x)
      hi = mid - 1;
    else
      lo = mid + 1, ans = mid;
  }
  return ans;
}
int main() {
  cin >> m;
  int l = get(m), r = get2(m);
  if (!ok) {
    cout << 0 << endl;
    return 0;
  }
  cout << r - l + 1 << endl;
  for (int i = l; i <= r; i++) cout << i << ' ';
  cout << endl;
  return 0;
}

```