#include <bits/stdc++.h>
using namespace std;
pair<int, int> adds[100002];
int asz;
int Cnt5(int num) {
  int ret = 0;
  while (num % 5 == 0 && num != 0) {
    ret++;
    num /= 5;
  }
  return ret;
}
int main() {
  int tar;
  scanf("%d", &tar);
  int curcnt = 0;
  for (int i = 0; i <= 100000; i++) {
    curcnt += Cnt5(i * 5);
    if (curcnt > tar) {
      printf("0\n");
      return 0;
    }
    if (curcnt == tar) {
      printf("5\n");
      for (int j = 5 * i; j < 5 * (i + 1); j++) {
        if (j != 5 * i) printf(" ");
        printf("%d", j);
      }
      printf("\n");
      return 0;
    }
  }
}

### Prompt

Generate 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;
pair<int, int> adds[100002];
int asz;
int Cnt5(int num) {
  int ret = 0;
  while (num % 5 == 0 && num != 0) {
    ret++;
    num /= 5;
  }
  return ret;
}
int main() {
  int tar;
  scanf("%d", &tar);
  int curcnt = 0;
  for (int i = 0; i <= 100000; i++) {
    curcnt += Cnt5(i * 5);
    if (curcnt > tar) {
      printf("0\n");
      return 0;
    }
    if (curcnt == tar) {
      printf("5\n");
      for (int j = 5 * i; j < 5 * (i + 1); j++) {
        if (j != 5 * i) printf(" ");
        printf("%d", j);
      }
      printf("\n");
      return 0;
    }
  }
}

```