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.StringTokenizer;

public class CF_1602a{
    public static final void main(String[] args){
        Kattio io= new Kattio();
        int t= io.getInt();
        while(t-->0){
            String s= io.getWord();
            int len= s.length();
            String ans= s;
            if(len>1){
                char first= s.charAt(0), last= s.charAt(len-1);
                if(first!=last){
                    ans= new StringBuilder().append(s.substring(0, len-1)).append(last=='a'?'b':'a').toString();
                }
            }
            io.println(ans);
        }
        io.close();
    }

    static void err(String format, Object... args){
        System.err.println(String.format(format, args));
    }    
    static class Kattio extends PrintWriter {
        private BufferedReader r;
        private String line, token;
        private StringTokenizer st;
    
        public Kattio(){this(System.in);}
        public Kattio(InputStream i){
        super(new BufferedOutputStream(System.out));
            r= new BufferedReader(new InputStreamReader(i));
        }
        public Kattio(InputStream i, OutputStream o){
        super(new BufferedOutputStream(o));
            r= new BufferedReader(new InputStreamReader(i));
        }
        public boolean isLineEnd(){
            return !st.hasMoreTokens();
        }
        public boolean hasMoreTokens(){
            return peekToken()!=null;
        }
        public int getInt(){
            return Integer.parseInt(nextToken());
        }
        public double getDouble(){ 
            return Double.parseDouble(nextToken());
        }
        public long getLong(){
            return Long.parseLong(nextToken());
        }
        public String getWord(){
            return nextToken();
        }
        private String peekToken(){
            if(token==null) try {
                while(st==null || !st.hasMoreTokens()) {
                    line= r.readLine();
                    if(line==null) return null;
                    st= new StringTokenizer(line);
                }
                token= st.nextToken();
            }catch(IOException e){}
            return token;
        }
        private String nextToken() {
            String ans= peekToken();
            token= null;
            return ans;
        }
    }    
}