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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  
  public class Main {
  
  	private static PrintWriter out = new PrintWriter(System.out);
  	
  	static int get(char[]s) {
  		int a = 0;
  		for(int i = 0;i < s.length - 1;i++) {
  			if(s[i] == 'a' && s[i+1]=='b')a++;
  			else if(s[i]=='b' &&s[i+1]=='a') a--;
  		}
  		return a;
  	}
  
  	public static void main(String[] args) throws IOException {
  		Scanner sc = new Scanner(System.in);
  		
  		int tc = sc.nextInt();
  		
  		outer:while(tc-->0) {
  			char[] s = sc.next().toCharArray();
  			
  			int a = get(s);
  			if(a==0) {
  				out.println(s);
  				continue;
  			}	
  			
  			for(int i=0;i<s.length;i++) {
  				s[i] = s[i]=='a'?'b':'a';
  				a = get(s);
  				if(a==0) {
  					out.println(s);
  					continue outer;
  				}
  				s[i] = s[i]=='a'?'b':'a';
  			}
  		}
  		
  		
  		out.flush();
  	}
  
  	private static class Scanner {
  		public BufferedReader reader;
  		public StringTokenizer st;
  
  		public Scanner(InputStream stream) {
  			reader = new BufferedReader(new InputStreamReader(stream));
  			st = null;
  		}
  
  		public String next() {
  			while (st == null || !st.hasMoreTokens()) {
  				try {
  					String line = reader.readLine();
  					if (line == null)
  						return null;
  					st = new StringTokenizer(line);
  				} catch (Exception e) {
  					throw (new RuntimeException());
  				}
  			}
  			return st.nextToken();
  		}
  
  		public int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  		public long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		public double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  
  }
  