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.util.*;
  import java.io.*;
  import java.math.*;
  
  public class Coder {
  
    static StringBuffer str=new StringBuffer();
    static String st;
    static char c[];
    static int n;
  
    static void solve(){
      int ab=0,ba=0;
      for(int i=0;i<n-1;i++){
        if(c[i]=='a' && c[i+1]=='b') ab++;
        else if(c[i]=='b' && c[i+1]=='a') ba++;
      }
      if(ab == ba){
        str.append(st).append("\n");
        return;
      }else if(ab > ba){
        int d=ab-ba;
        if(c[n-1]!='a'){
          c[n-1]='a';
          d--;
        }
        if(d>0){
          d--;
          c[0]='b';
        }
      }else{
        int d=ba-ab;
        if(c[n-1]!='b'){
          c[n-1]='b';
          d--;
        }
        if(d>0){
          d--;
          c[0]='a';
        }
      }
      for(int i=0;i<n;i++) str.append(c[i]);
        str.append("\n");
  
    }
  
    public static void main(String[] args) throws java.lang.Exception {
      BufferedReader bf;
      PrintWriter pw;
      boolean lenv=false;
      if(lenv){
        bf = new BufferedReader(
                            new FileReader("input.txt"));
        pw=new PrintWriter(new
              BufferedWriter(new FileWriter("output.txt")));
      }else{
        bf = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(new OutputStreamWriter(System.out));
      }
      int q = Integer.parseInt(bf.readLine().trim());
      while (q-- > 0) {
        st=bf.readLine().trim();
        c=st.toCharArray();
        n=c.length;
        solve();
      }
      pw.print(str);
      pw.flush();
      // System.out.print(str);
    }
  }