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 Aqueous {
  		
  		static MyScanner sc = new MyScanner();
  		
  		public static void main(String[] args) {
  
  			int t = sc.nextInt();
  			
  			while(t-->0) {
  				
  				 String s = sc.next();
  				 
  				 int ab =0;
  				 int ba = 0;
  				 
  				 for(int i =0; i<s.length()-1; i++) {
  					 if(s.charAt(i)=='a' && s.charAt(i+1)=='b') {
  						 ab++;
  					 }
  					 else if(s.charAt(i)=='b' && s.charAt(i+1)=='a') {
  						 ba++;
  					 }
  				 }
  				 
  				 if(ab==ba) {
  					 System.out.println(s);
  				 }
  				 else if(ab>ba) {
  					 StringBuffer sb  = new StringBuffer();
  					 sb.append(s);
  					 for(int i = 0; i<s.length(); i++) {
  						 if(s.charAt(i)=='a') {
  							 sb.setCharAt(i, 'b');
  							 break;
  						 }
  					 }
  					 
  					 System.out.println(sb.toString());
  				 }
  				 else {
  					 StringBuffer sb  = new StringBuffer();
  					 sb.append(s);
  					 for(int i = 0; i<s.length(); i++) {
  						 if(s.charAt(i)=='b') {
  							 sb.setCharAt(i, 'a');
  							 break;
  						 }
  					 }
  					 
  					 System.out.println(sb.toString());
  				 }
  			}
  		}
  		
  		
  		static void ruffleSort(int a[]) {
  
  	          int n = a.length;
  
  	          Random r = new Random();
  
  	          for (int i = 0; i < n; i++) {
  
  	              int oi = r.nextInt(n);
  	              int temp = a[oi];
  	              a[oi] = a[i];
  	              a[i] = temp;
  	          }
  
  	          Arrays.sort(a);
  	      }
  				
  		static void ruffleSort(long a[]) {
  
  	          int n = a.length;
  
  	          Random r = new Random();
  
  	          for (int i = 0; i < n; i++) {
  	              int oi = r.nextInt(n);
  
  	              long temp = a[oi];
  	              a[oi] = a[i];
  	              a[i] = temp;
  	          }
  
  	          Arrays.sort(a);
  	      }
  
  		
  	    public static class MyScanner {
  	        BufferedReader br;
  	        StringTokenizer st;
  
  	        public MyScanner() {
  	            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;
  	        }
  
  	    }
  	}
  