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.StringTokenizer;
  
  public class CF_1602a{
      public static final void main(String[] args){
          Kattio io= new Kattio();
          int t= io.getInt();
          while(t-->0){
              String s= io.getWord();
              int len= s.length();
              String ans= s;
              if(len>1){
                  char first= s.charAt(0), last= s.charAt(len-1);
                  if(first!=last){
                      ans= new StringBuilder().append(s.substring(0, len-1)).append(last=='a'?'b':'a').toString();
                  }
              }
              io.println(ans);
          }
          io.close();
      }
  
      static void err(String format, Object... args){
          System.err.println(String.format(format, args));
      }    
      static class Kattio extends PrintWriter {
          private BufferedReader r;
          private String line, token;
          private StringTokenizer st;
      
          public Kattio(){this(System.in);}
          public Kattio(InputStream i){
          super(new BufferedOutputStream(System.out));
              r= new BufferedReader(new InputStreamReader(i));
          }
          public Kattio(InputStream i, OutputStream o){
          super(new BufferedOutputStream(o));
              r= new BufferedReader(new InputStreamReader(i));
          }
          public boolean isLineEnd(){
              return !st.hasMoreTokens();
          }
          public boolean hasMoreTokens(){
              return peekToken()!=null;
          }
          public int getInt(){
              return Integer.parseInt(nextToken());
          }
          public double getDouble(){ 
              return Double.parseDouble(nextToken());
          }
          public long getLong(){
              return Long.parseLong(nextToken());
          }
          public String getWord(){
              return nextToken();
          }
          private String peekToken(){
              if(token==null) try {
                  while(st==null || !st.hasMoreTokens()) {
                      line= r.readLine();
                      if(line==null) return null;
                      st= new StringTokenizer(line);
                  }
                  token= st.nextToken();
              }catch(IOException e){}
              return token;
          }
          private String nextToken() {
              String ans= peekToken();
              token= null;
              return ans;
          }
      }    
  }