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.
1
abbabababbabababbababbbabbbbabbbabaabaabba
 4
b
aabbbabaa
abbb
abbaab
 1
aaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbab

abbabababbabababbababbbabbbbabbbabaabaabba
 b
aabbbabaa
bbbb
bbbaab
 baaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbab

import java.io.*;
import java.util.*;
import static java.lang.Math.*;

public class A{
	public static void main(String[] args) throws IOException {
		// br = new BufferedReader(new FileReader(".in"));
		// out = new PrintWriter(new FileWriter(".out"));
		//new Thread(null, new (), "peepee", 1<<28).start();
		int t    =readInt();
		while(t-->0) {
			int ab = 0, ba  = 0;
			char[] s  =  read().toCharArray();
			for (int  i  =  0; i <s.length-1;  i++) {
				if (s[i] == 'a' && s[i+1]  ==  'b') ab++;
				else if (s[i] == 'b' && s[i+1] == 'a') ba++;
			}
			if (ba  ==  ab) out.println(s);
			else {
				int  i  = 0;
				if  (s[i] == 'a') s[i] = 'b';
				else  s[i]  = 'a';
				out.println(s);
			}
		}
		out.close();
	}
	
	/* Stupid things to try if stuck:
	 * n=1, expand bsearch range
	 * brute force small patterns
	 * submit stupid intuition
	 */
	
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
	static StringTokenizer st = new StringTokenizer("");
	static String read() throws IOException{return st.hasMoreTokens() ? st.nextToken():(st = new StringTokenizer(br.readLine())).nextToken();}
	static int readInt() throws IOException{return Integer.parseInt(read());}
	static long readLong() throws IOException{return Long.parseLong(read());}
	static double readDouble() throws IOException{return Double.parseDouble(read());}
}