Problem: 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.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.lang.*;
public class BinaryPeriod{	
	public static void main(String args[]){ 
		try{
	 		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
			String[] inp = br.readLine().trim().split("\\s+");
			int n = Integer.parseInt(inp[0]);
			for(int i =0 ; i< n ; i++){
                inp = br.readLine().trim().split("");
                int flag = 0;
                for(int op = 1 ; op<inp.length ; op++){
                    if(!inp[op].equals(inp[op-1])){
                        flag=1;
                        //System.out.println("herere"+i);
                        break;
                    }
                }
                if(flag==1)
                    for(int op = 0 ; op<inp.length ; op++)
                        System.out.print("01");
                else
                    for(int op = 0 ; op<inp.length ; op++)
                        System.out.print(inp[op]);                    
                System.out.println("");
			}		
		}
		catch(Exception E){}
	}
}