Hackerrank. Day 3: If-Else Statements!

Problem Statement
Welcome to Day 3! Check out a review of if-else statements here, or just jump right into the problem.
Using "if-else" you can perform decision making in your code. See the flowchart below (taken from wikipedia):
This problem will test your knowledge on "if-else" statements.
Given an integer N as input, check the following:
  • If N is odd, print "Weird".
  • If N is even and, in between the range of 2 and 5(inclusive), print "Not Weird".
  • If N is even and, in between the range of 6 and 20(inclusive), print "Weird".
  • If N is even and N>20, print "Not Weird".
We have given you partially completed code in the editor, complete it to solve the problem.
Input Format
There is a single line of input: integer N.
Constraints 
1≤N≤100
Output Format
Print "Weird" if the number is weird. Otherwise, print "Not Weird". Do not print the quotation marks.
Sample Input 1
3
Sample Output 1
Weird

Explanation 
N=3, is odd hence the its a Weird Number.
Sample Input 2
24
Sample Output 2
Not Weird

Explanation 
N=24, is >20 hence its not a Weird Number.
Current Buffer (saved locally, editable)     

 Solution for above problem :

   import java.io.*;
    import java.util.*;
    import java.text.*;
    import java.math.*;
    import java.util.regex.*;

    public class Solution {

        public static void main(String[] args) {

            Scanner sc=new Scanner(System.in);
            int n=sc.nextInt();          
            String ans="";
            if(n%2==1)ans = "Weird";
            else
            {
               //Complete the code
                if (n >= 2 && n <= 5)
                {
                    ans = "Not Weird";
                }
                else if (n >= 6 && n <= 20)
                {
                    ans = "Weird";
                }
                else
                {
                    ans = "Not Weird";
                }  
               
            }
            System.out.println(ans);
           
        }
    }

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