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.File;
  import java.io.FileInputStream;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintStream;
  import java.io.PrintWriter;
  import java.time.Duration;
  import java.time.Instant;
  import java.util.ArrayList;
  import java.util.StringTokenizer;
  import static java.lang.Math.min;
  import static java.lang.Math.max;
  import static java.lang.Math.abs;
  
  @SuppressWarnings("unused")
  public class A {
  	static boolean DEBUG = false;
  
  	public static void main(String[] args) throws IOException {
  		Instant start = Instant.now();
  		if (args.length == 2) {
  			System.setIn(new FileInputStream(new File("D:\\program\\javaCPEclipse\\CodeForces\\src\\input.txt")));
  			// System.setOut(new PrintStream(new File("output.txt")));
  			System.setErr(new PrintStream(new File("D:\\program\\javaCPEclipse\\CodeForces\\src\\error.txt")));
  			DEBUG = true;
  		}
  		Reader fs = new Reader();
  		PrintWriter pw = new PrintWriter(System.out);
  		int t = fs.nextInt();
  		while (t-- > 0) {
  			char a[] = fs.next().toCharArray();
  			int n = a.length;
  			int ab = 0, ba = 0;
  			for (int i = 1; i < n; i++) {
  				if (a[i - 1] == 'a' && a[i] == 'b') {
  					ab++;
  				}
  				if (a[i - 1] == 'b' && a[i] == 'a') {
  					ba++;
  				}
  			}
  			if (ab == ba) {
  				for (char x : a)
  					pw.print(x);
  				pw.println();
  				continue;
  			}
  			if (ab > ba) {
  				if (a[0] == 'a' && a[1] == 'b') {
  					a[0] = 'b';
  				} else if (a[n - 1] == 'b' && a[n - 2] == 'b') {
  					a[n - 1] = 'a';
  				} else if (a[n - 1] == 'a' && a[n - 2] == 'a') {
  					a[n - 2] = 'b';
  				} else if (a[1] == 'b' && a[0] == 'b') {
  					a[1] = 'a';
  				} else if (a[1] == 'a' && a[0] == 'a') {
  					a[0] = 'b';
  				} else if (a[n-2] == 'a' && a[n-1] == 'b') {
  					a[n-2] = 'b';
  				}
  				for (char x : a)
  					pw.print(x);
  				pw.println();
  				continue;
  			}
  			if (ba > ab) {
  				if (a[0] == 'b' && a[1] == 'a') {
  					a[0] = 'a';
  				} else if (a[n - 1] == 'a' && a[n - 2] == 'a') {
  					a[n - 1] = 'b';
  				} else if (a[n - 1] == 'b' && a[n - 2] == 'b') {
  					a[n - 2] = 'a';
  				} else if (a[1] == 'a' && a[0] == 'a') {
  					a[1] = 'b';
  				} else if (a[1] == 'b' && a[0] == 'b') {
  					a[0] = 'a';
  				} else if (a[n-2] == 'b' && a[n-1] == 'a') {
  					a[n-2] = 'a';
  				}
  
  				for (char x : a)
  					pw.print(x);
  				pw.println();
  				continue;
  			}
  		}
  		Instant end = Instant.now();
  		if (DEBUG) {
  			pw.println(Duration.between(start, end));
  		}
  		pw.close();
  	}
  
  	public static void print(long a, long b, long c, PrintWriter pw) {
  		pw.println(a + " " + b + " " + c);
  		return;
  	}
  
  	static class Reader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public Reader() {
  			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;
  		}
  
  		int[] readArray(int n) {
  			int a[] = new int[n];
  			for (int i = 0; i < n; i++)
  				a[i] = nextInt();
  			return a;
  		}
  
  		int[][] read2Array(int n, int m) {
  			int a[][] = new int[n][m];
  			for (int i = 0; i < n; i++) {
  				for (int j = 0; j < m; j++) {
  					a[i][j] = nextInt();
  				}
  			}
  			return a;
  		}
  	}
  }