Trailing zeroes in factorial

import java.util.*;

public class Main {
    public static void main(String[] args) {
       
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        for(int t=0;t<T;t++){
            int N = sc.nextInt();
            int count=0;
            for(;N>=5;N/=N/){
                N=N/5;
                count= count+N/5;
            }
            System.out.println(count);
        }
    }
}
//Explanation:
//The number of trailing zeroes in the factorial of a number nn depends on the number of times 55 and 22 appear in its prime factors, as each pair of 2 and 5 contributes a trailing zero.
//Since there are always more factors of 2 than 5 in factorials, the count of trailing zeros is determined by the number of times 5 appears in the factorization.
//You continue dividing by higher powers of 5 until the quotient becomes zero. This approach efficiently counts the factors of 5 in n!.