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.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  
  public class a {
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer("");
  
          int tc = Integer.parseInt(br.readLine());
  
          for (int tt = 0; tt < tc; tt++) {
              String s = br.readLine();
              int cnt1 = 0;
              int cnt2 = 0;
              char[] answer = new char[s.length()];
              for (int i = 0; i < s.length(); i++) answer[i] = s.charAt(i);
  
              for (int i = 0; i < s.length() - 1; i++) {
                  if (s.charAt(i) == 'a' && s.charAt(i + 1) == 'b') {
                      cnt1++;
                  } else if (s.charAt(i) == 'b' && s.charAt(i + 1) == 'a') {
                      cnt2++;
                  }
              }
  
              if (cnt1 == cnt2) {
                  System.out.println(s);
                  continue;
              }
  
              if(cnt1 == cnt2 +1){ //ab > ba
                  for(int i=0;i<s.length();i++){
                      if(answer[i]=='a') {
                          answer[i]='b';
                          break;
                      }
                  }
              } else if(cnt1 +1 == cnt2){ //ab < ba
                  for(int i=0;i<s.length();i++){
                      if(answer[i]=='b') {
                          answer[i]='a';
                          break;
                      }
                  }
              }
  
              for(int i=0;i<s.length();i++) System.out.print(answer[i]);
              System.out.println();
          }
      }
  
      private static int gcd(int a, int b) {
          if (a % b == 0) return b;
          else return gcd(b, a % b);
      }
  }