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.BufferedWriter;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.OutputStreamWriter;
  import java.io.PrintWriter;
  import java.io.StreamTokenizer;
  import java.math.BigInteger;
  import java.util.ArrayList;
  import java.util.HashMap;
  import java.util.HashSet;
  import java.util.Iterator;
  import java.util.PriorityQueue;
  import java.util.Scanner;
  import java.util.TreeSet;
  
  public class Main {
  	StreamTokenizer in=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
  	PrintWriter out=new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
  	int n,t,r,b,i;
  	void run() throws IOException{
  		Scanner in=new Scanner(System.in);
  		t=in();
  		while(t-->0)
  		{
  			String str=ins();
  			n=str.length();
  			if(str.charAt(0)==str.charAt(n-1)) {
  				out.println(str);
  			}else {
  				out.println(str.charAt(n-1)+str.substring(1));
  			}
  		}
  		out.flush();
  	}
  	public static void main(String[]args) throws IOException {
  		new Main().run();
  	}
  	public int in() throws IOException {
  		in.nextToken();
  		return(int)in.nval;
  	}
  	public long inl() throws IOException {
  		in.nextToken();
  		return(long)in.nval;
  	}
  	public double ind() throws IOException {
  		in.nextToken();
  		return in.nval;
  	}
  	public String ins() throws IOException {
  		in.nextToken();
  		return in.sval;
  	}
  }
  
  