#include <bits/stdc++.h>
using namespace std;
const int N = 120000;
vector<long long> G[N];
int solve(int x) {
  int ans = 0;
  while (x) {
    ans += x = x / 5;
  }
  return ans;
}
void init() {
  for (int i = 0; i <= 450000; i++) {
    int k = solve(i);
    G[k].push_back(i);
  }
}
int main() {
  init();
  int n;
  scanf("%d", &n);
  int m = G[n].size();
  printf("%d\n", m);
  for (int i = 0; i < m; i++) {
    printf("%d", G[n][i]);
    if (i < m - 1)
      printf(" ");
    else
      printf("\n");
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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 N = 120000;
vector<long long> G[N];
int solve(int x) {
  int ans = 0;
  while (x) {
    ans += x = x / 5;
  }
  return ans;
}
void init() {
  for (int i = 0; i <= 450000; i++) {
    int k = solve(i);
    G[k].push_back(i);
  }
}
int main() {
  init();
  int n;
  scanf("%d", &n);
  int m = G[n].size();
  printf("%d\n", m);
  for (int i = 0; i < m; i++) {
    printf("%d", G[n][i]);
    if (i < m - 1)
      printf(" ");
    else
      printf("\n");
  }
  return 0;
}

```