#include <bits/stdc++.h>
using namespace std;
int i, num[1000010], cnt[100010];
int main() {
  int n;
  scanf("%d", &n);
  num[0] = 0;
  for (i = 1; i < 1000010; i++) {
    if (i < 100010) cnt[i] = 0;
    num[i] = 1 + ((i % 5 == 0) ? num[i / 5] : 0);
  }
  for (i = 1; i < 1000010; i++) {
    num[i] += num[i - 1];
    if (num[i] < 100010) cnt[num[i]] = i;
  }
  if (cnt[n] == 0)
    printf("0\n");
  else {
    printf("%d\n", 5);
    for (i = 5 * cnt[n]; i < 5 * cnt[n] + 5; i++) printf("%d ", i);
    printf("\n");
  }
}

### Prompt

Please create a solution in CPP 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 i, num[1000010], cnt[100010];
int main() {
  int n;
  scanf("%d", &n);
  num[0] = 0;
  for (i = 1; i < 1000010; i++) {
    if (i < 100010) cnt[i] = 0;
    num[i] = 1 + ((i % 5 == 0) ? num[i / 5] : 0);
  }
  for (i = 1; i < 1000010; i++) {
    num[i] += num[i - 1];
    if (num[i] < 100010) cnt[num[i]] = i;
  }
  if (cnt[n] == 0)
    printf("0\n");
  else {
    printf("%d\n", 5);
    for (i = 5 * cnt[n]; i < 5 * cnt[n] + 5; i++) printf("%d ", i);
    printf("\n");
  }
}

```