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: // letsgoooooooooooooooooooooooooooooooo
  
  import java.util.*;
  import java.io.*;
   public class Solution{
       
      static int MOD=1000000007;
      static PrintWriter pw;
      static FastReader sc;
      
      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());}
          public char nextChar() throws IOException {return next().charAt(0);}
          String nextLine(){
              String str = "";
              try {
                  str = br.readLine();
              }
              catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
      
      static void Solve() throws Exception{
          String str=sc.next();
          int [] arr=check(str);
          if(arr[2]==1){
              pw.println(str);
              return;
          }
          if(arr[0]>arr[1]){
              int diff=arr[0]-arr[1];
              StringBuilder sb=new StringBuilder(str);
              if(diff==1 && sb.charAt(0)=='a'){
                  sb.replace(0,1,"b");
                  pw.println(sb);
                  return;
              }
              for(int i=1;i<str.length()-1;i++){
                  if(sb.charAt(i)=='a' && sb.charAt(i-1)!='a' && sb.charAt(i+1)=='b'){
                      sb.replace(i,i+1,"b");
                      diff--;
                      if(diff==0){
                          pw.println(sb);
                          return;
                      }
                  }
              }
          }else{
              int diff=arr[1]-arr[0];
              StringBuilder sb=new StringBuilder(str);
              if(diff==1 && sb.charAt(0)=='b'){
                  sb.replace(0,1,"a");
                  pw.println(sb);
                  return;
              }
              for(int i=1;i<str.length();i++){
                  if(sb.charAt(i)=='b' && sb.charAt(i-1)!='b'  && sb.charAt(i+1)=='a'){
                      sb.replace(i,i+1,"a");
                      diff--;
                      if(diff==0){
                          pw.println(sb);
                          return;
                      }
                  }
              }
          }
  
  
  }
  
  public static int[] check(String str){
      int [] arr= new int [3];
      int ab=0,ba=0;
      for(int i=0;i<str.length()-1;i++){
          if(str.charAt(i)=='a' && str.charAt(i+1)=='b'){
              ab++;
              arr[0]++;
          }
          if( str.charAt(i)=='b' && str.charAt(i+1)=='a' ){
              ba++;
              arr[1]++;
          }
      }
      if(ab==ba){
          arr[2]=1;
      }
      return arr;
  
  }
  
     public static void main(String[] args) throws Exception{
          try {
              System.setIn(new FileInputStream("input.txt"));
              System.setOut(new PrintStream(new FileOutputStream("output.txt")));
          } catch (Exception e) {
              System.err.println("Error");
          }
          sc= new FastReader();
          pw = new PrintWriter(System.out);
  
     
          int t=1;
          t=sc.nextInt();
  
          for(int ii=1;ii<=t;ii++) {
              Solve();
          }
         
          pw.flush();
      }
  }