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.BufferedOutputStream;
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.math.BigInteger;
  import java.sql.Time;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.Collections;
  import java.util.Comparator;
  import java.util.Date;
  import java.util.Deque;
  import java.util.HashMap;
  import java.util.HashSet;
  import java.util.LinkedList;
  import java.util.Map;
  import java.util.Scanner;
  import java.util.Stack;
  import java.util.StringTokenizer;
  import java.util.TreeMap;
  
  public class Practice1 {
  	
  	
  	public static void main(String[] args) {
  		PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
  		// out.print();
  		//out.println();
  		FastReader sc=new FastReader();
  		int t=sc.nextInt();
  		while(t-->0) {
  			String str=sc.nextLine();
  			int n=str.length();
  			if(str.length()==1||str.charAt(0)==str.charAt(n-1)) {
  				out.println(str);
  			}else {
  			   out.println("a"+str.substring(1,n-1)+"a");
  			}
  		}
  		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;
  	     }
  	 }
  }
  