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 A1606{
  		static FastScanner fs = null;
  		public static void main(String[] args) {
  			fs = new FastScanner();
  			PrintWriter out = new PrintWriter(System.out);
  			int t = fs.nextInt();
  			while(t-->0){
  				String s = fs.next();
  				char ch[] = s.toCharArray();
  				int n = s.length();
  				int count = 0;
  				char prev = ch[0];
  				for(int i=0;i<n;i++){
  					if(ch[i]!=prev){
  						count+=1;
  					}
  					prev = ch[i];
  				}
  				if(count%2!=0){
  					char ch1 = ch[0];
  					ch[n-1] = ch1;
  				}
  				String s2 = "";
  				for(int i=0;i<n;i++){
  					s2+=ch[i];
  				}
  				out.println(s2);
  			}
  			out.println();
  			out.close();
  		}
  		static void sort(int[] a) {
  			ArrayList<Integer> l=new ArrayList<>();
  			for (int i:a) l.add(i);
  			Collections.sort(l);
  			for (int i=0; i<a.length; i++) a[i]=l.get(i);
  		}
  		static class FastScanner {
  			BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  			StringTokenizer st=new StringTokenizer("");
  			String next() {
  				while (!st.hasMoreTokens())
  					try {
  						st=new StringTokenizer(br.readLine());
  					} catch (IOException e) {
  						e.printStackTrace();
  					}
  				return st.nextToken();
  			}
  			
  			int nextInt() {
  				return Integer.parseInt(next());
  			}
  			int[] readArray(int n) {
  				int[] a=new int[n];
  				for (int i=0; i<n; i++) a[i]=nextInt();
  				return a;
  			}
  			long nextLong() {
  				return Long.parseLong(next());
  			}
  		}
  	}