#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 5, M = 1e5 + 5, OO = 0x3f3f3f3f;
int m;
long long findE(long long x, long long p) {
  long long res = 0;
  while (x) {
    res += x / p;
    x /= p;
  }
  return res;
}
bool ok(int x) { return findE(x, 5) < m; }
int binarySearch() {
  int lo = 0, med, hi = 1e6;
  while (lo < hi) {
    med = (lo + hi) >> 1;
    if (ok(med))
      lo = med + 1;
    else
      hi = med;
  }
  return lo;
}
int main() {
  scanf("%d", &m);
  int ans = binarySearch();
  if (findE(ans, 5) == m) {
    puts("5");
    for (int i = 0; i < 5; ++i) printf("%d ", ans + i);
    puts("");
  } else
    puts("0");
  return 0;
}

### Prompt

Please provide a cpp coded solution 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>
using namespace std;
const int N = 1e3 + 5, M = 1e5 + 5, OO = 0x3f3f3f3f;
int m;
long long findE(long long x, long long p) {
  long long res = 0;
  while (x) {
    res += x / p;
    x /= p;
  }
  return res;
}
bool ok(int x) { return findE(x, 5) < m; }
int binarySearch() {
  int lo = 0, med, hi = 1e6;
  while (lo < hi) {
    med = (lo + hi) >> 1;
    if (ok(med))
      lo = med + 1;
    else
      hi = med;
  }
  return lo;
}
int main() {
  scanf("%d", &m);
  int ans = binarySearch();
  if (findE(ans, 5) == m) {
    puts("5");
    for (int i = 0; i < 5; ++i) printf("%d ", ans + i);
    puts("");
  } else
    puts("0");
  return 0;
}

```