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:
//Codeforces B. Binary Period 1100

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

public class CodeforcesBinaryPeriod
{
	private static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

	private static StringBuffer ans = new StringBuffer();

	private static void solution(String st) {
		int i;
		boolean flag = false;
		char[] seq = st.toCharArray();
		for (i = 1; i < seq.length; i++) {
			if (seq[i] != seq[0]) {
				flag = !flag;
				break;
			}
		}
		StringBuffer str = new StringBuffer(st);
		if (flag && st.length() > 2) {
			for (i = 1; i < str.length(); i++) {
				if (str.charAt(i) == str.charAt(i - 1)) {
					if (str.charAt(i) == '0') {
						str.insert(i,  '1');
					} else {
						str.insert(i, '0');
					}
					i++;
				}
			}
		}
		ans.append(str + "\n");
	}

	public static void main(String[] args)throws IOException {
		try{
			final long startTime = System.currentTimeMillis();
			int test;
			// int[] arr;
			String st;
			for (test = Integer.parseInt(br.readLine()); test > 0; test--) {
				// arr = Arrays.stream(br.readLine().split(" ")).mapToInt(Integer::valueOf).toArray();
				st = br.readLine();
				solution(st);
			}
			System.out.print(ans);
			final long finishTime = System.currentTimeMillis();
			// System.out.println("Time:" + (finishTime - startTime) + " ms");
		} catch (Exception e) {
			e.printStackTrace();
		}
	}
}