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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6;
int meme[MAXN + 3];
int rec(int a) {
  if (meme[a] != -1) {
    return meme[a];
  }
  if (a == 0) {
    return 0;
  } else {
    int ret = a / 5 + rec(a / 5);
    meme[a] = ret;
    return ret;
  }
}
int main(int argc, char* argv[]) {
  int m;
  scanf("%d", &m);
  fill(&meme[0], &meme[0] + sizeof(meme) / sizeof(int), -1);
  vector<int> ans;
  for (int i = 1; i < 1e6; i++) {
    if (rec(i) == m) {
      ans.push_back(i);
    }
  }
  printf("%d\n", ans.size());
  for (auto i : ans) {
    printf("%d ", i);
  }
  printf("\n");
  return 0;
}