#include <bits/stdc++.h>
using namespace std;
string i2s(int x) {
  stringstream ss;
  ss << x;
  return ss.str();
}
int s2i(string str) {
  istringstream ss(str);
  int nro;
  ss >> nro;
  return nro;
}
int n;
int NRO5(int nro) {
  int cnt = 0;
  while (nro % 5 == 0) {
    cnt++;
    nro /= 5;
  }
  return cnt;
}
int main() {
  int A[500005];
  map<int, vector<int> > mapa;
  map<int, vector<int> >::iterator it;
  for (int i = 1; i < 500005; i++) {
    A[i] = A[i - 1] + NRO5(i);
    mapa[A[i]].push_back(i);
  }
  scanf("%d\n", &n);
  if (mapa[n].size() == 0) {
    putchar('0');
    putchar('\n');
  } else {
    printf("%d\n", (int)mapa[n].size());
    for (int i = 0; i < mapa[n].size(); i++) {
      printf("%d%c", mapa[n][i], i == mapa[n].size() - 1 ? 10 : 32);
    }
  }
  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>
using namespace std;
string i2s(int x) {
  stringstream ss;
  ss << x;
  return ss.str();
}
int s2i(string str) {
  istringstream ss(str);
  int nro;
  ss >> nro;
  return nro;
}
int n;
int NRO5(int nro) {
  int cnt = 0;
  while (nro % 5 == 0) {
    cnt++;
    nro /= 5;
  }
  return cnt;
}
int main() {
  int A[500005];
  map<int, vector<int> > mapa;
  map<int, vector<int> >::iterator it;
  for (int i = 1; i < 500005; i++) {
    A[i] = A[i - 1] + NRO5(i);
    mapa[A[i]].push_back(i);
  }
  scanf("%d\n", &n);
  if (mapa[n].size() == 0) {
    putchar('0');
    putchar('\n');
  } else {
    printf("%d\n", (int)mapa[n].size());
    for (int i = 0; i < mapa[n].size(); i++) {
      printf("%d%c", mapa[n][i], i == mapa[n].size() - 1 ? 10 : 32);
    }
  }
  return 0;
}

```