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 binaryperiod {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int cases = scan.nextInt();
		for (int i = 0; i < cases; i++) {
			//StringTokenizer st= new StringTokenizer(scan.nextLine());
			String s = scan.next();
			if (s.matches("0+")|| s.matches("1+"))
				System.out.println(s);
			else {
				char[] chars = new char[s.length()*2];
				Arrays.fill(chars, '0');
				for (int j=0;j<chars.length;j+=2)
					chars[j]='1';
					String ans = new String(chars);
					System.out.println(ans);
			}
			
		}
		
	}
}
		
