#include <bits/stdc++.h>
const int inf = (1 << 30) - 10;
using namespace std;
inline int get_int() {
  int r = 0;
  char c;
  while ((c = getchar()) != ' ' && c != '\n') r = r * 10 + c - '0';
  return r;
}
inline void out(int x) {
  if (x > 10) {
    out(x / 10);
  }
  putchar(x % 10 + '0');
  putchar('\n');
}
void put(int n) {
  printf("5\n");
  for (int i = 0; i < 5; i++) {
    printf("%d ", i + n);
  }
  printf("\n");
}
int main() {
  int n, l, x;
  scanf("%d", &n);
  l = 0;
  for (int i = 0;; i += 5) {
    x = i;
    while (x) {
      if (x % 5 == 0) {
        l++;
        x /= 5;
      } else
        break;
    }
    if (l == n) {
      put(i);
      break;
    }
    if (l > n) {
      printf("0\n");
      break;
    }
  }
  return 0;
}

### Prompt

Your task is to create 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>
const int inf = (1 << 30) - 10;
using namespace std;
inline int get_int() {
  int r = 0;
  char c;
  while ((c = getchar()) != ' ' && c != '\n') r = r * 10 + c - '0';
  return r;
}
inline void out(int x) {
  if (x > 10) {
    out(x / 10);
  }
  putchar(x % 10 + '0');
  putchar('\n');
}
void put(int n) {
  printf("5\n");
  for (int i = 0; i < 5; i++) {
    printf("%d ", i + n);
  }
  printf("\n");
}
int main() {
  int n, l, x;
  scanf("%d", &n);
  l = 0;
  for (int i = 0;; i += 5) {
    x = i;
    while (x) {
      if (x % 5 == 0) {
        l++;
        x /= 5;
      } else
        break;
    }
    if (l == n) {
      put(i);
      break;
    }
    if (l > n) {
      printf("0\n");
      break;
    }
  }
  return 0;
}

```