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.*;
  import java.util.*;
  
  public class Main {
      public static void main(String[] args) throws IOException {
          int t = nextInt();
          for (int i = 0; i < t; i++) {
              String s = nextToken();
              int count1 = 0;
              int count2 = 0;
              for (int j = 0; j < s.length() - 1; j++) {
                  if (s.charAt(j) == 'a' && s.charAt(j + 1) == 'b') count1++;
                  if (s.charAt(j) == 'b' && s.charAt(j + 1) == 'a') count2++;
              }
              if (count1 == count2) out.println(s);
              else {
                  for (int j = 0; j < s.length(); j++) {
                      String s1 = s.substring(0, j);
                      if (s.charAt(j) == 'a') {
                          s1 += 'b';
                      } else {
                          s1 += 'a';
                      }
                      s1 += s.substring(j + 1);
                      int counta = 0;
                      int countb = 0;
                      for (int k = 0; k < s1.length() - 1; k++) {
                          if (s1.charAt(k) == 'a' && s1.charAt(k + 1) == 'b') counta++;
                          if (s1.charAt(k) == 'b' && s1.charAt(k + 1) == 'a') countb++;
                      }
                      if (counta == countb) {
                          out.println(s1);
                          break;
                      }
                  }
              }
          }
          out.close();
      }
  
      static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      static PrintWriter out = new PrintWriter(System.out);
      static StringTokenizer in = new StringTokenizer("");
  
  
      public static boolean hasNext() throws IOException {
          if (in.hasMoreTokens()) return true;
          String s;
          while ((s = br.readLine()) != null) {
              in = new StringTokenizer(s);
              if (in.hasMoreTokens()) return true;
          }
          return false;
      }
  
      public static String nextToken() throws IOException {
          while (!in.hasMoreTokens()) {
              in = new StringTokenizer(br.readLine());
          }
          return in.nextToken();
      }
  
      public static int nextInt() throws IOException {
          return Integer.parseInt(nextToken());
      }
  
      public static double nextDouble() throws IOException {
          return Double.parseDouble(nextToken());
      }
  
      public static long nextLong() throws IOException {
          return Long.parseLong(nextToken());
      }
  }
  
  class Ayo implements Comparable<Ayo> {
      long r;
      long l;
  
      public Ayo(long r, long l) {
          this.r = r;
          this.l = l;
      }
  
      @Override
      public int compareTo(Ayo o) {
          return -Long.compare(o.r, r);
      }
  }