For this Question: Given an integer N, return the number of trailing zeroes in N!. 
Where N!=N*(N-1)*(N-2)....1Input contains a single integer N.

Constraints:-
1<= N <=1000000000000000Print the number of zeros in factorial(N)Sample Input
5

Sample Output
1

Explanation:-
5! = 120
number of zeroes = 1

Sample Input;
11

Sample output:
2, I have written this Solution Code: import java.io.*;
import java.util.*;



class Main {
	static long findTrailingZeros(long n)
    {
        long count = 0;
  	  for (long i = 5; n / i >= 1; i *= 5)
            count += n / i;
        return count;
    }
	static long fact(long num)
	{
		long sum=1;
		for(int i=1;i<num;i++)
		{
			sum+=sum*i;
		}
		return sum;
	}
	public static void main (String[] args)throws IOException {
    BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
	long num=Long.parseLong(bf.readLine());
System.out.println(findTrailingZeros(num));
	}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?