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 Exception {
          FastReader input=new FastReader();
          PrintWriter out=new PrintWriter(System.out);
          int T=input.nextInt();
          while (T-- > 0)
          {
              String n=input.next();
              
              if(n.charAt(0)==n.charAt(n.length()-1))
               out.println(n);
              else if(n.charAt(0)=='a') out.println('b'+n.substring(1));
              else if(n.charAt(0)=='b') out.println('a'+n.substring(1));
          }
          out.close();
      }
      
      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;
          }
      }
  }