#include <bits/stdc++.h>
int getint() {
  int r = 0;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; '0' <= c && c <= '9'; c = getchar()) r = r * 10 - '0' + c;
  return r;
}
int n, z = 0, two = 0, five = 0;
int min(int a, int b) { return a < b ? a : b; }
int dst[200000], ptr = 0;
int main() {
  n = getint();
  int i = 1, cnt, tmp;
  while (z <= n) {
    tmp = i;
    while (tmp % 2 == 0) ++two, tmp /= 2;
    while (tmp % 5 == 0) ++five, tmp /= 5;
    cnt = min(two, five);
    z += cnt;
    two -= cnt;
    five -= cnt;
    if (z == n) dst[ptr++] = i;
    ++i;
  }
  printf("%d\n", ptr);
  for (int i = 0; i < ptr; ++i) printf("%d ", dst[i]);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
int getint() {
  int r = 0;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; '0' <= c && c <= '9'; c = getchar()) r = r * 10 - '0' + c;
  return r;
}
int n, z = 0, two = 0, five = 0;
int min(int a, int b) { return a < b ? a : b; }
int dst[200000], ptr = 0;
int main() {
  n = getint();
  int i = 1, cnt, tmp;
  while (z <= n) {
    tmp = i;
    while (tmp % 2 == 0) ++two, tmp /= 2;
    while (tmp % 5 == 0) ++five, tmp /= 5;
    cnt = min(two, five);
    z += cnt;
    two -= cnt;
    five -= cnt;
    if (z == n) dst[ptr++] = i;
    ++i;
  }
  printf("%d\n", ptr);
  for (int i = 0; i < ptr; ++i) printf("%d ", dst[i]);
  return 0;
}

```