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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class B {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintStream out = System.out;

        int t = Integer.parseInt(br.readLine());

        for (int i = 0; i < t; i++) {
            String curr = br.readLine();
            boolean allTheSame = true;
            for (int j = 1; j < curr.length(); j++) {
                if (curr.charAt(j) != curr.charAt(j-1)) allTheSame = false;
            }
            if (allTheSame) out.println(curr);
            else {
                StringBuilder ans = new StringBuilder();
                char first = curr.charAt(0);
                for (int j = 0; j < 2*curr.length(); j++) {
                    ans.append(first);
                    if (first == '0') first = '1';
                    else first = '0';
                }
                out.println(ans.toString());
            }
        }


        System.out.flush();
    }
}