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 MyClass {
    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());
      for(int p=0;p<t;p++)
      {
          String s=br.readLine();
          boolean b=false;
          for(int i=0;i<s.length()-1;i++)
          {
              if(s.charAt(i)!=s.charAt(i+1))
              {
                  b=true;
                  break;
              }
          }
          if(b)
          {
              if(s.charAt(0)=='0')
              {
                  for(int i=0;i<2*s.length();i++)
                  {
                      if(i%2==0)
                      pw.print(0);
                      else
                      pw.print(1);
                  }
                  pw.println();
              }
              else
              {
                  for(int i=0;i<2*s.length();i++)
                  {
                      if(i%2==0)
                      pw.print(1);
                      else
                      pw.print(0);
                  }
                  pw.println();
              }
          }
          else
          pw.println(s);
      }
      pw.flush();
      pw.close();
    }
}