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: /*Everything is Hard 
   * Before Easy 
   * Jai Mata Dii 
   */ 
   
  import java.util.*;
  import java.io.*; 
    
  public class Main {
  	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; }} 
  	static long mod = (long)(1e9+7); 
  //	 static long mod = 998244353; 
  //	 static Scanner sc = new Scanner(System.in); 
  	static FastReader sc = new FastReader(); 
  	static PrintWriter out = new PrintWriter(System.out);
  	public static void main (String[] args) { 
  		int ttt = 1;
  		ttt = sc.nextInt();
  		z :for(int tc=1;tc<=ttt;tc++){
  			String S = sc.next();
  			char s[] = S.toCharArray();
  			int n = s.length;
  			int ab = 0, ba = 0;
  			for(int i=1;i<n;i++) {
  				if(s[i] == 'a' && s[i-1] == 'b') {
  					ba++;
  				}
  				if(s[i] == 'b' && s[i-1] == 'a') {
  					ab++;
  				}
  			}
  			if(ab == ba) {
  				out.write(S+"\n");
  				continue;
  			}
  			if(ab>ba) {
  				for(int i=n-1;i>0;i--) {
  					if(s[i]=='b' && s[i-1]=='b') {
  						s[i] = 'a';
  						break;
  					}
  					else if(s[i] == 'b' && s[i-1] == 'a') {
  						s[i] = 'a';
  						break;
  					}
  				}
  			}
  			else {
  				for(int i=n-1;i>0;i--) {
  					if(s[i]=='a' && s[i-1]=='a') {
  						s[i] = 'b';
  						break;
  					}
  					else if(s[i] == 'a' && s[i-1] == 'b') {
  						s[i] = 'b';
  						break;
  					}
  				}
  			}
  			for(char c : s) {
  				out.write(c+"");
  			}
  			out.write("\n");
  		}
  		out.close();
  	}
  	static long pow(long a, long b){long ret = 1;while(b>0){if(b%2 == 0){a = (a*a)%mod;b /= 2;}else{ret = (ret*a)%mod;b--;}}return ret%mod;}
  	static long gcd(long a,long b){if(b==0) return  a; return gcd(b,a%b); } 
  	private static void sort(int[] a) {List<Integer> k = new ArrayList<>();for(int val : a) k.add(val);Collections.sort(k);for(int i=0;i<a.length;i++) a[i] = k.get(i);} 
  	private static void ini(List<Integer>[] tre2){for(int i=0;i<tre2.length;i++){tre2[i] = new ArrayList<>();}} 
  	private static void init(List<int[]>[] tre2){for(int i=0;i<tre2.length;i++){tre2[i] = new ArrayList<>();}} 
  	private static void sort(long[] a) {List<Long> k = new ArrayList<>();for(long val : a) k.add(val);Collections.sort(k);for(int i=0;i<a.length;i++) a[i] = k.get(i);} 
  }