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.*;
public class binarytree {
    public static void main(String[] quesgs) {
        Scanner sc = new Scanner(System.in);
        
        String testcase = sc.nextLine();     
        int n = Integer.parseInt(testcase);
        
        String[] arr = new String[n];
        
        for(int j = 0; j<n; j++){
            String a = sc.nextLine();
            char[] brr = a.toCharArray();
            int ab = 0;
            int ba = 0;
            int size = brr.length;
            
            for(int k = 0; k<size-1; k++){
                if(brr[k]=='a' && brr[k+1]=='b'){
                    ab++;
                }
                else if(brr[k]=='b' && brr[k+1]=='a'){
                    ba++;
                }
            }
            
            if(brr[0]==brr[size-1])
            {
               arr[j]=String.copyValueOf(brr);
            }
            else{
                if(brr[0]=='a'){
                    brr[size-1]='a';
                }
                else if(brr[0]=='b'){
                    brr[size-1]='b';
                }
                arr[j]=String.copyValueOf(brr);
            }
        }
        

        for(int i = 0; i<n; i++){
            System.out.println(arr[i]);
        }
        
    }
}