Let's say string s has period k if s_i = s_{i + k} for all i from 1 to |s| - k (|s| means length of string s) and k is the minimum positive integer with this property.

Some examples of a period: for s="0101" the period is k=2, for s="0000" the period is k=1, for s="010" the period is k=2, for s="0011" the period is k=4.

You are given string t consisting only of 0's and 1's and you need to find such string s that:

  1. String s consists only of 0's and 1's; 
  2. The length of s doesn't exceed 2 ⋅ |t|; 
  3. String t is a subsequence of string s; 
  4. String s has smallest possible period among all strings that meet conditions 1—3. 



Let us recall that t is a subsequence of s if t can be derived from s by deleting zero or more elements (any) without changing the order of the remaining elements. For example, t="011" is a subsequence of s="10101".

Input

The first line contains single integer T (1 ≤ T ≤ 100) — the number of test cases.

Next T lines contain test cases — one per line. Each line contains string t (1 ≤ |t| ≤ 100) consisting only of 0's and 1's.

Output

Print one string for each test case — string s you needed to find. If there are multiple solutions print any one of them.

Example

Input


4
00
01
111
110


Output


00
01
11111
1010

Note

In the first and second test cases, s = t since it's already one of the optimal solutions. Answers have periods equal to 1 and 2, respectively.

In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string s. String s has period equal to 1.
// package Round_86;

import java.io.*;
import java.util.*;

public class Binary_Period {  
    public static void main(String[] args) {
        FastReader in = new FastReader();
        
        int numTrials = in.nextInt();
        while(numTrials --> 0) {
            StringBuffer str = new StringBuffer(in.next());
            
            int len = str.length();
            
            StringBuilder outStr = new StringBuilder();
            if(str.indexOf("0") == -1) {
                for(int i = 0; i < len * 2; i++) {
                    outStr.append("1");
                }
            }
            else if(str.indexOf("1") == -1) {
                for(int i = 0; i < len * 2; i++) {
                    outStr.append("0");
                }
            }
            else {
                boolean next1 = true;
                if(str.substring(0, 1).equals("0")) {
                    next1 = false;
                }
                
                for(int i = 0; i < str.length() * 2; i++) {                   
                    if(next1) {
                        outStr.append("1");
                    }
                    else {
                        outStr.append("0");
                    }
                    
                    next1 = !next1;
                }
            }
            
            System.out.println(outStr);
        }
    }
    
    static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        }
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
}
