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.*;
public class Main{
    
    public static void main(String args[]) throws IOException{
       Scanner sc=new Scanner(System.in);
       int i,j,k,n,t,c1,c2;
       t=sc.nextInt();
       sc.nextLine();
       for(i=0;i<t;i++)
       { c1=0;c2=0;
        String s=sc.nextLine();
        String z="";
        n=s.length();
        for(j=0;j<n;j++)
        { if(s.charAt(j)=='0')
            c1++;
            else
            c2++;
        }
        if(c1==0||c2==0)
        System.out.println(s);
        else
        { k=0;for(j=0;j<n-1;j++)
            { if(s.charAt(j)=='0')
                { z+="0";
                    if(s.charAt(j+1)=='1')
                    k=1;
                    else
                    {z+="1";
                     k=3;
                    }
                }
                else
                { z+="1";
                    if(s.charAt(j+1)=='0')
                    k=1;
                    else
                    {z+="0";
                    k=2;
                    }
                }
            }
           /* if(k==1||k==0)
            z+=Character.toString(s.charAt(n-1));
            else if(k==2)
            z+=Character.toString(s.charAt(n-1));*/
            if(z.length()%2!=0)
            z+=Character.toString(s.charAt(n-1));
            else 
            {if(s.charAt(j)=='0')
                { z+="0";
                //System.out.println(z+" "+s);
                    if(z.length()-2>=0&&z.charAt(z.length()-2)=='1')
                    z+="1";
                }
                else
                { z+="1";
                    if(z.length()-2>=0&&z.charAt(z.length()-2)=='0')
                    z+="0";
                }
                
            }
             System.out.println(z);
        }
       }
        
       
            
        
    }
}