#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
long long int n, a[10000005], b[10000005];
vector<long long int> v;
void solve() {
  long long int i, j, sum = 0, q;
  for (i = 1; i < 10000005; i++) {
    q = i;
    if (i == 1) {
      a[i] = 0;
      b[i] = 0;
    } else {
      a[i] = a[i - 1];
      b[i] = b[i - 1];
    }
    while (q > 0 && q % 2 == 0) {
      a[i]++;
      q /= 2;
    }
    while (q > 0 && q % 5 == 0) {
      b[i]++;
      q /= 5;
    }
  }
  for (i = 0; i < 10000005; i++) {
    if (min(a[i], b[i]) == n) {
      v.push_back(i);
      sum++;
    }
  }
  printf("%I64d\n", sum);
  for (i = 0; i < sum; i++) {
    printf("%I64d ", v[i]);
  }
  return;
}
int main() {
  scanf("%I64d", &n);
  solve();
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
#pragma warning(disable : 4996)
using namespace std;
long long int n, a[10000005], b[10000005];
vector<long long int> v;
void solve() {
  long long int i, j, sum = 0, q;
  for (i = 1; i < 10000005; i++) {
    q = i;
    if (i == 1) {
      a[i] = 0;
      b[i] = 0;
    } else {
      a[i] = a[i - 1];
      b[i] = b[i - 1];
    }
    while (q > 0 && q % 2 == 0) {
      a[i]++;
      q /= 2;
    }
    while (q > 0 && q % 5 == 0) {
      b[i]++;
      q /= 5;
    }
  }
  for (i = 0; i < 10000005; i++) {
    if (min(a[i], b[i]) == n) {
      v.push_back(i);
      sum++;
    }
  }
  printf("%I64d\n", sum);
  for (i = 0; i < sum; i++) {
    printf("%I64d ", v[i]);
  }
  return;
}
int main() {
  scanf("%I64d", &n);
  solve();
  return 0;
}

```