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.*;
public class BinaryPeriod {
    public static void main(String args[])
    {
        Scanner in = new Scanner(System.in);
        int n = Integer.parseInt(in.nextLine());
        for(int i = 0; i < n; i++)
        {
            String str = in.nextLine();
            if(str.indexOf("0") < 0 || str.indexOf("1") < 0)
            {
                System.out.println(str);
            }
            else
            {
                String empty = "";
                for(int k = 0; k < str.length() - 1; k++)
                {
                    empty += str.substring(k, k +1);
                    if(str.substring(k, k + 1).equals(str.substring(k + 1, k + 2)))
                    {
                        if(str.substring(k, k + 1).equals("0"))
                        {
                            empty += "1";
                        }
                        else
                        {
                            empty += "0";
                        }
                    }
                }
                empty += str.substring(str.length() - 1, str.length());
                System.out.println(empty);
            }
        }
    }
}