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.Scanner;
  import java.util.Arrays;
  import java.util.List;
  import java.util.LinkedList;
  import java.util.ArrayList;
  import java.util.Set;
  import java.util.HashSet;
  import java.util.Map;
  import java.util.HashMap;
  
  import java.io.BufferedWriter;
  import java.io.OutputStreamWriter;
  
  public class A {
  	public static void main(String... args) {
  		try {
  			Scanner sc = new Scanner(System.in);
  			BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
  
  			int t = sc.nextInt();
  			for (; t > 0; t--) {
  				String s = sc.next();
  
  				int ab = count("ab", s); int ba = count("ba", s);
  				StringBuilder sb = new StringBuilder(s);
  
  				if (ab > ba) {
  					sb.replace(0, 1, "b");
  				} else if (ab < ba) {
  					sb.replace(sb.length()-1, sb.length(), "b");
  				}
  
  				System.out.println(sb);
  			}
  
  			sc.close();
  			writer.close();
  		} catch (Exception e) {
  			e.printStackTrace();
  		}
  	}
  
  	private static int count(String p, String s) {
  		int pos = 0;
  		int cnt = 0;
  
  		int next = 0;
  		while ((next = s.indexOf(p, pos)) >= 0) {
  			cnt++; pos = next+1;
  		}
  
  		return cnt;
  	}
  }
  