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.*; 
import java.lang.Math; 
public class Main
{   public static void main(String[] args)
{ 
    Scanner in=new Scanner(System.in); 
    int i,n,temp,f=0,j,m;
    n=in.nextInt();
    for(i=0;i<n;i++) 
    {
        f=0;m=0;
        String t,k="",s="";
        t=in.next();
        char ch[]=t.toCharArray();
        for(j=1;j<t.length();j++)
        {
            if(ch[0]==ch[j])
            {
                f++;
            }
        }
        if(f==t.length()-1)
        {
            System.out.println(t);     
/*00
01
11111
1010*/
        }
        else
        {
        //    if(ch[0]=='1')
            //{
               k+="01";
            //}
          //  else
        //    {
              //  k+="10";
            //}
            
            for(j=0;j<t.length();j++)
            {
                s+=k;
            }
            System.out.println(s);
        }
        
        
        
    }
}
}