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.io.*;
import java.util.*;

public class CF1342B extends PrintWriter {
	CF1342B() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1342B o = new CF1342B(); o.main(); o.flush();
	}

	void main() {
		int t = sc.nextInt();
		while (t-- > 0) {
			byte[] aa = sc.next().getBytes();
			int n = aa.length;
			boolean one = true;
			for (int i = 1; i < n; i++)
				if (aa[i] != aa[0]) {
					one = false;
					break;
				}
			if (one)
				println(new String(aa));
			else {
				byte[] bb = new byte[n * 2];
				for (int i = 0; i < n * 2; i++)
					bb[i] = (byte) ('0' + i % 2);
				println(new String(bb));
			}
		}
	}
}