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.
import java.util.*;
import java.io.*;
public class BinaryPeriod
{
	public static void main(String[] args)throws Exception
	{
	    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	    PrintWriter pw=new PrintWriter(System.out);
	    int T=Integer.parseInt(br.readLine());
	    while(T-->0)
	    {
	        int one=-1;
	        int zero=-1;
	        String t=(br.readLine()).trim();
	        for(int i=0;i<t.length();i++)
	        {
	            if(t.charAt(i)=='1')
	            {
	                one=1;
	            }
	            else if(t.charAt(i)=='0')
	            {
	                zero=1;
	            }
	        }
	        if((one==-1 && zero==1)||(one==1 && zero==-1))
	        {
	            if(t.length()>1)
	            {
	                pw.println(t);
	            }
	            else
	            {
	                String x=t;
	                x+=t;
	                pw.println(x);
	            }
	        }
	        else
	        {
	            String s="";
	            for(int i=0;i<t.length();i++)
	            {
	                s+="10";
	            }
	            pw.println(s);
	        }
	        
	    }
	    pw.flush();
	}
}
