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.*;
  
  public class A_1606 {
  	public static void main(String[] args) throws Exception {
  		Scanner sc = new Scanner(System.in);
  		PrintWriter pw = new PrintWriter(System.out);
  		
  		int t = sc.nextInt();
  		while(t-->0) {
  			String s = sc.next();
  			int n = s.length();
  			int cnt = 1;
  			int flip = 0;
  			for(int i = 1; i < n; i++) {
  				if(s.charAt(i) != s.charAt(i - 1)) {
  					cnt = 1;
  					flip++;
  				} else {
  					cnt++;
  				}
  			}
  			if(flip % 2 == 0)
  				pw.println(s);
  			else {
  				pw.println((s.charAt(0) == 'a' ? 'b' : 'a') + (s.substring(1)));
  			}
  		}
  		
  		pw.flush();
  	}
  
  	public static class Scanner {
  		StringTokenizer st;
  		BufferedReader br;
  
  		public Scanner(InputStream system) {
  			br = new BufferedReader(new InputStreamReader(system));
  		}
  
  		public Scanner(String file) throws Exception {
  			br = new BufferedReader(new FileReader(file));
  		}
  
  		public String next() throws IOException {
  			while (st == null || !st.hasMoreTokens())
  				st = new StringTokenizer(br.readLine());
  			return st.nextToken();
  		}
  
  		public String nextLine() throws IOException {
  			return br.readLine();
  		}
  
  		public int nextInt() throws IOException {
  			return Integer.parseInt(next());
  		}
  
  		public double nextDouble() throws IOException {
  			return Double.parseDouble(next());
  		}
  
  		public char nextChar() throws IOException {
  			return next().charAt(0);
  		}
  
  		public long nextLong() throws IOException {
  			return Long.parseLong(next());
  		}
  
  		public int[] nextIntArray(int n) throws IOException {
  			int[] array = new int[n];
  			for (int i = 0; i < n; i++)
  				array[i] = nextInt();
  			return array;
  		}
  
  		public Integer[] nextIntegerArray(int n) throws IOException {
  			Integer[] array = new Integer[n];
  			for (int i = 0; i < n; i++)
  				array[i] = new Integer(nextInt());
  			return array;
  		}
  
  		public long[] nextLongArray(int n) throws IOException {
  			long[] array = new long[n];
  			for (int i = 0; i < n; i++)
  				array[i] = nextLong();
  			return array;
  		}
  
  		public double[] nextDoubleArray(int n) throws IOException {
  			double[] array = new double[n];
  			for (int i = 0; i < n; i++)
  				array[i] = nextDouble();
  			return array;
  		}
  
  		public static int[] shuffle(int[] a) {
  			int n = a.length;
  			Random rand = new Random();
  			for (int i = 0; i < n; i++) {
  				int tmpIdx = rand.nextInt(n);
  				int tmp = a[i];
  				a[i] = a[tmpIdx];
  				a[tmpIdx] = tmp;
  			}
  			return a;
  		}
  
  		public boolean ready() throws IOException {
  			return br.ready();
  		}
  
  		public void waitForInput() throws InterruptedException {
  			Thread.sleep(3000);
  		}
  	}
  }
  