#include <bits/stdc++.h>
using namespace std;
int q;
int n;
inline int zeros(int n) {
  int ret = 0;
  while (n > 0) {
    n /= 5;
    ret += n;
  }
  return ret;
}
int explore(int start, int end) {
  int mid;
  int count;
  int t;
  if (start > end) {
    return -1;
  }
  mid = (start + end) / 2;
  count = zeros(mid);
  if (count == n) {
    while (zeros(mid) == n) {
      mid--;
    }
    return ++mid;
  }
  if (count < n) {
    return explore(mid + 1, end);
  } else {
    return explore(start, mid - 1);
  }
}
int main() {
  int t;
  int ans;
  scanf("%d", &n);
  ans = explore(1, INT_MAX);
  if (ans != -1) {
    printf("%d\n", 5);
    printf("%d %d %d %d %d\n", ans, ans + 1, ans + 2, ans + 3, ans + 4);
  } else {
    printf("0\n");
  }
}

### 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;
int q;
int n;
inline int zeros(int n) {
  int ret = 0;
  while (n > 0) {
    n /= 5;
    ret += n;
  }
  return ret;
}
int explore(int start, int end) {
  int mid;
  int count;
  int t;
  if (start > end) {
    return -1;
  }
  mid = (start + end) / 2;
  count = zeros(mid);
  if (count == n) {
    while (zeros(mid) == n) {
      mid--;
    }
    return ++mid;
  }
  if (count < n) {
    return explore(mid + 1, end);
  } else {
    return explore(start, mid - 1);
  }
}
int main() {
  int t;
  int ans;
  scanf("%d", &n);
  ans = explore(1, INT_MAX);
  if (ans != -1) {
    printf("%d\n", 5);
    printf("%d %d %d %d %d\n", ans, ans + 1, ans + 2, ans + 3, ans + 4);
  } else {
    printf("0\n");
  }
}

```