description: You are given a string s of length n consisting of characters a and/or b.
  
  Let \operatorname{AB}(s) be the number of occurrences of string ab in s as a substring. Analogically, \operatorname{BA}(s) is the number of occurrences of ba in s as a substring.
  
  In one step, you can choose any index i and replace s_i with character a or b.
  
  What is the minimum number of steps you need to make to achieve \operatorname{AB}(s) = \operatorname{BA}(s)?
  
  Reminder:
  
  The number of occurrences of string d in s as substring is the number of indices i (1 ≤ i ≤ |s| - |d| + 1) such that substring s_i s_{i + 1} ... s_{i + |d| - 1} is equal to d. For example, \operatorname{AB}(aabbbabaa) = 2 since there are two indices i: i = 2 where aabbbabaa and i = 6 where aabbbabaa.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.
  
  The first and only line of each test case contains a single string s (1 ≤ |s| ≤ 100, where |s| is the length of the string s), consisting only of characters a and/or b.
  
  Output
  
  For each test case, print the resulting string s with \operatorname{AB}(s) = \operatorname{BA}(s) you'll get making the minimum number of steps.
  
  If there are multiple answers, print any of them.
  
  Example
  
  Input
  
  
  4
  b
  aabbbabaa
  abbb
  abbaab
  
  
  Output
  
  
  b
  aabbbabaa
  bbbb
  abbaaa
  
  Note
  
  In the first test case, both \operatorname{AB}(s) = 0 and \operatorname{BA}(s) = 0 (there are no occurrences of ab (ba) in b), so can leave s untouched.
  
  In the second test case, \operatorname{AB}(s) = 2 and \operatorname{BA}(s) = 2, so you can leave s untouched. 
  
  In the third test case, \operatorname{AB}(s) = 1 and \operatorname{BA}(s) = 0. For example, we can change s_1 to b and make both values zero.
  
  In the fourth test case, \operatorname{AB}(s) = 2 and \operatorname{BA}(s) = 1. For example, we can change s_6 to a and make both values equal to 1.
solution: import java.util.*;
  public class Ol{
  	public static void main(String[] args) {
  		Scanner scan = new Scanner(System.in);
  		int n = scan.nextInt();
  		for (int i = 0; i < n; i++){
  			String s = scan.next();
  			if(s.length() == 1 || (s.length() == 2 && (s.charAt(0))==(s.charAt(1)))){
  				System.out.println(s);
  			}
  			else if(s.length() == 2){
  				System.out.println("aa");
  			}
  			else{
  				int f = 0;
  				int se = 0;
  				for(int j = 0; j < s.length() - 2; j++){
                      if((s.substring(j, j+2)).equals("ab")){
                      	f++;
                      	
                      }
                      if((s.substring(j, j+2)).equals("ba")){
                      	se++;
                      	
                      }
  				}
  				if((s.substring(s.length()-2)).equals("ab")){
  					f++;
  					
  				}
  				if((s.substring(s.length()-2)).equals("ba")){
  					se++;
  					
  				}
  				if(f == se){
  					System.out.println(s);
  					
  				}
  				else{
  					if(f > se){
  						if((s.substring(0, 2)).equals("ab")){
  							s = "b" + "" + s.substring(1);
  							
  						}
  						else if((s.substring(s.length()-2)).equals("ab")){
  							s = s.substring(0, s.length() - 2) + "" + "aa";
  							
  						}
  						else if((s.substring(0, 2)).equals("aa")){
  							s="b" + "" +s.substring(1);
  						}
  						else if((s.substring(s.length() - 2)).equals("bb")){
  							s=s.substring(0, s.length() - 1) + "" + "a";
  						}
  					}
  					else{
  						if((s.substring(0, 2)).equals("ba")){
  							s = "aa" + "" + s.substring(2);
  							
  						}
  						else if((s.substring(s.length()-2)).equals("ba")){
  							s = s.substring(0, s.length()-2) + "" + "bb";
  							
  						}
  						else if((s.substring(0, 2)).equals("bb")){
  							s = "a" + "" + s.substring(1);
  						}
  						else if((s.substring(s.length() - 2)).equals("aa")){
  							s = s.substring(0, s.length() - 1) + "" + "b";
  						}
  					}
  					System.out.println(s);
  				}
  			}
  			
  		}
  	}
  }