#include <bits/stdc++.h>
using namespace std;
const int LIM = 1e7;
int two[LIM], five[LIM], ten[LIM];
int cnt[LIM];
inline int mini(int a, int b) { return a < b ? a : b; }
int main(void) {
  int m;
  scanf("%d", &m);
  vector<int> ans;
  two[2] = 1;
  for (int i = 3; i < LIM; i++) {
    int tmp = i;
    two[i] = two[i - 1];
    while (!(tmp % 2)) two[i]++, tmp /= 2;
    five[i] = five[i - 1];
    while (!(tmp % 5)) five[i]++, tmp /= 5;
    ten[i] = ten[i - 1];
    while (!(tmp % 10)) ten[i]++, tmp /= 10;
    cnt[i] = mini(two[i], five[i]) + ten[i];
    if (cnt[i] == m)
      ans.push_back(i);
    else if (cnt[i] > m)
      break;
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) {
    if (i) putchar(' ');
    printf("%d", ans[i]);
  }
  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;
const int LIM = 1e7;
int two[LIM], five[LIM], ten[LIM];
int cnt[LIM];
inline int mini(int a, int b) { return a < b ? a : b; }
int main(void) {
  int m;
  scanf("%d", &m);
  vector<int> ans;
  two[2] = 1;
  for (int i = 3; i < LIM; i++) {
    int tmp = i;
    two[i] = two[i - 1];
    while (!(tmp % 2)) two[i]++, tmp /= 2;
    five[i] = five[i - 1];
    while (!(tmp % 5)) five[i]++, tmp /= 5;
    ten[i] = ten[i - 1];
    while (!(tmp % 10)) ten[i]++, tmp /= 10;
    cnt[i] = mini(two[i], five[i]) + ten[i];
    if (cnt[i] == m)
      ans.push_back(i);
    else if (cnt[i] > m)
      break;
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) {
    if (i) putchar(' ');
    printf("%d", ans[i]);
  }
  return 0;
}

```