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.BufferedWriter;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.OutputStreamWriter;
  import java.util.StringTokenizer;
  
  /**
   *
   * @author eslam
   */
  public class JavaApplication778 {
  
      static class FastReader {
  
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() {
              br = new BufferedReader(
                      new InputStreamReader(System.in));
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  
      public static void main(String[] args) throws IOException {
          BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
          FastReader input = new FastReader();
          int t = input.nextInt();
          for (int i = 0; i < t; i++) {
              String w = input.next();
              int ab = 0;
              int ba = 0;
              for (int j = 0; j < w.length(); j++) {
                  if (w.charAt(j) == 'a') {
                      while (j < w.length()) {
                          if (w.charAt(j) == 'b') {
                              ab++;
                              j--;
                              break;
                          }
                          j++;
                      }
                  } else if (w.charAt(j) == 'b') {
                      while (j < w.length()) {
                          if (w.charAt(j) == 'a') {
                              ba++;
                              j--;
                              break;
                          }
                          j++;
                      }
                  }
  
              }
              if (ab == ba) {
                  log.write(w + "\n");
              } else if (ab < ba) {
                  log.write(w.substring(0, w.length()-1)+"b"+"\n");
              } else {
                  log.write(w.substring(0, w.length()-1)+"a"+"\n");
              }
          }
          log.flush();
      }
  
  }
  