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.lang.reflect.Array;
  import java.text.SimpleDateFormat;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.Calendar;
  import java.util.Collections;
  import java.util.Deque;
  import java.util.HashMap;
  import java.util.LinkedList;
  import java.util.Map;
  import java.util.PriorityQueue;
  import java.util.Queue;
  import java.util.Scanner;
  import java.util.Stack;
  import java.util.Map.Entry;
  import javax.naming.ldap.SortControl;
  public class Main
  {
      public static void main(String[] args)
      {
          Scanner in=new Scanner(System.in);
          int t=in.nextInt();
          while(t-->0)
          {
          	String zifu=in.next();
          	char z[]=zifu.toCharArray();
          	int ab=0;
          	int ba=0;
          	for(int i=0;i<z.length;i++)
          	{
          		if(i+1<z.length)
          		{
          			if(z[i]=='a'&&z[i+1]=='b')
          			{
          				ab++;
          			}
          			if(z[i]=='b'&&z[i+1]=='a')
          			{
          				ba++;
          			}
          		}
          	}
          	if(ab!=ba)
          	{
          		if(ab>ba)
          		{
          			for(int i=z.length-2;i>=0;i--)
          			{
          				if(z[i]=='a'&&z[i+1]=='b')
          				{
          					z[i+1]='a';
          					ab--;
          					if(ab==ba)break;
          				}
          				else
          				{
          					if(z[i]=='b'&&z[i+1]=='b')
          					{
          						z[i+1]='a';
          						ba++;
          						if(ab==ba)break;
          					}
          				}
          			}
          		}
          		else
          		{
          			for(int i=z.length-2;i>=0;i--)
          			{
          				if(z[i]=='b'&&z[i+1]=='a')
          				{
          					z[i+1]='b';
          					ba--;
          					if(ab==ba)break;
          				}
          				else
          				{
          					if(z[i]=='a'&&z[i+1]=='a')
          					{
          						z[i+1]='b';
          						ab++;
          						if(ab==ba)break;
          					}
          				}
          			}
          		}
          	}
          	for(int i=0;i<z.length;i++)
          	{
          		System.out.print(z[i]);
          	}
          	System.out.println();
          }
      }
  }
  