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 Main {
  
  //	static boolean[] prime = new boolean[10000000];
  	final static long mod = 1000000007;
  
  	public static void main(String[] args) {
  //		sieve();
  		InputReader in = new InputReader(System.in);
  		PrintWriter out = new PrintWriter(System.out);
  
  		int t = in.nextInt();
  		while (t-- > 0) {
  			String str = in.next();
  			if(str.charAt(0) == str.charAt(str.length()-1)) {
  				out.println(str);
  			}else {
  				out.print(str.charAt(str.length() - 1));
  				out.println(str.substring(1));
  			}
  		}
  
  		out.flush();
  	}
  
  	static void rewrite(Integer[] a) {
  		HashMap<Integer, Integer> map = new HashMap<>();
  		for(int i : a) {
  			map.put(i, map.getOrDefault(i, 0)+1);
  		}
  		for(int i = 0; i < a.length; i++) {
  			a[i] = map.get(a[i]);
  		}
  	}
  	
  	static long gcd(long a, long b) {
  		if (a % b == 0) {
  			return b;
  		} else {
  			return gcd(b, a % b);
  		}
  	}
  
  	static void reverseArray(int[] a) {
  		for (int i = 0; i < (a.length >> 1); i++) {
  			int temp = a[i];
  			a[i] = a[a.length - 1 - i];
  			a[a.length - 1 - i] = temp;
  		}
  	}
  
  	static Integer[] intInput(int n, InputReader in) {
  		Integer[] a = new Integer[n];
  		for (int i = 0; i < a.length; i++)
  			a[i] = in.nextInt();
  		return a;
  	}
  
  	static Long[] longInput(int n, InputReader in) {
  		Long[] a = new Long[n];
  		for (int i = 0; i < a.length; i++)
  			a[i] = in.nextLong();
  		return a;
  	}
  
  	static String[] strInput(int n, InputReader in) {
  		String[] a = new String[n];
  		for (int i = 0; i < a.length; i++)
  			a[i] = in.next();
  		return a;
  	}
  
  //	static void sieve() {
  //		for (int i = 2; i * i < prime.length; i++) {
  //			if (prime[i])
  //				continue;	
  //			for (int j = i * i; j < prime.length; j += i) {
  //				prime[j] = true;
  //			}
  //		}
  //	}
  
  }
  
  class Data {
  	int val;
  	int ind;
  	int i;
  
  	Data(int val, int ind, int i) {
  		this.val = Math.abs(val);
  		this.ind = ind;
  		this.i = i;
  	}
  }
  
  class compareVal implements Comparator<Data> {
  
  	@Override
  	public int compare(Data o1, Data o2) {
  //		return (o1.val - o2.val == 0 ? o1.ind - o2.ind : o1.val - o2.val);
  		return (o1.i - o2.i);
  	}
  
  }
  
  class compareInd implements Comparator<Data> {
  
  	@Override
  	public int compare(Data o1, Data o2) {
  		return o1.ind - o2.ind == 0 ? o1.val - o2.val : o1.ind - o2.ind;
  	}
  
  }
  
  class InputReader {
  	public BufferedReader reader;
  	public StringTokenizer tokenizer;
  
  	public InputReader(InputStream stream) {
  		reader = new BufferedReader(new InputStreamReader(stream), 32768);
  		tokenizer = null;
  	}
  
  	public String next() {
  		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
  			try {
  				tokenizer = new StringTokenizer(reader.readLine());
  			} catch (IOException e) {
  				throw new RuntimeException(e);
  			}
  		}
  		return tokenizer.nextToken();
  	}
  
  	public int nextInt() {
  		return Integer.parseInt(next());
  	}
  
  	long nextLong() {
  		return Long.parseLong(next());
  	}
  
  	double nextDouble() {
  		return Double.parseDouble(next());
  	}
  
  	String nextLine() {
  		String str = "";
  		try {
  			str = reader.readLine();
  		} catch (IOException e) {
  			e.printStackTrace();
  		}
  		return str;
  	}
  
  }