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.math.BigInteger;
  import java.util.*;
  import java.io.*;
  
  public class A_AB_Balance {
      public static void main (String[] args) {
          FastReader scan = new FastReader();
          int t = scan.nextInt();
          while (t-- > 0) {
              String str = scan.nextLine();
              int ab = 0, ba = 0;
              int n = str.length();
              for (int i=0; i<n-1; i++) {
                  if (str.charAt(i) == 'a' && str.charAt(i+1) == 'b') {
                      ab++;
                  }
                  if (str.charAt(i) == 'b' && str.charAt(i+1) == 'a') {
                      ba++;
                  }
              }
              if (ab == ba) {
                  System.out.println(str);
                  continue;
              }
              if (str.charAt(0) == 'a') {
                  System.out.println("b" + str.substring(1, n));
              }
              else {
                  System.out.println("a" + str.substring(1, n));
              }
          }
      }
      public static void inputArray (int[] arr, int n, FastReader scan) {
          for (int i=0; i<n; i++) {
              arr[i] = scan.nextInt();
          }
      }
      public static void printArray (int[] arr) {
          for (int i=0; i<arr.length; i++) {
              System.out.print(arr[i] + " ");
          }
          System.out.println();
      }
      public static boolean isPrime (int n) {
          if (n <= 1)
              return false;
          if (n <= 3)
              return true;
          if (n % 2 == 0 || n % 3 == 0)
              return false;
          for (int i = 5; i * i <= n; i = i + 6)
              if (n % i == 0 || n % (i + 2) == 0)
                  return false;
          return true;
      }
      public static int gcd (int a, int b) {
          BigInteger a1 = BigInteger.valueOf(a);
          BigInteger b1 = BigInteger.valueOf(b);
          return a1.gcd(b1).intValue();
      }
      public static int numberOfDigitsInNumber (int n) {
          return (int)Math.floor(Math.log10(n)) + 1;
      }
      public static boolean isPowerOfTwo (int x) {
          return x!=0 && ((x&(x-1)) == 0);   
      }
      public static String toBinaryString (int n) {
          return Integer.toString(n, 2);
          // for other bases change 2 to other bases
      }
      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;
          }
      }
  }
  
  // For IO using Files
  // public static void main throws IOException
  // BufferedReader br = new BufferedReader(new FileReader("input.txt"));
  // PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
  // String st[]=br.readLine().split(" ");
  // int n=Integer.parseInt(st[0]);
  // int m=Integer.parseInt(st[1]);
  // at the end
  // br.close(); pw.close();
  // for printing pw.println(), pw.print(), etc...
  // for arrays : int arr[] = Arrays.stream(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();