Problem: 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.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class Solution{
	public static void main(String[] args) throws Exception{
		Scanner sc = new Scanner(System.in);

		int T = sc.nextInt();

		while(T-- > 0){
			String t = sc.next();
			String s = "";

			int count0 = 0;

			for(int i = 0; i < t.length(); i++){
				if(t.charAt(i) == '0') count0++;
			}

			if(count0 == t.length() || count0 == 0){
				System.out.println(t);
			}
			else{
				for(int i = 0; i < 2 * t.length(); i++){
					if(i % 2 == 0) System.out.print("0");
					else System.out.print("1");
				}
				System.out.println();
			}
		}
	}
}