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.*;

public class problemB {
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int T = Integer.parseInt(input.nextLine());
		for (int t = 0; t < T; t++) {
			String str = input.nextLine();
			if (periodOne(str)) {
				System.out.println(str);
			} else {
				String ans = "";
				char curr = str.charAt(0);
				ans = ans + curr;
				for (int i = 1; i < str.length(); i++) {
					char nw = str.charAt(i);
					if (nw != ans.charAt(ans.length()-1)) {
						ans = ans + nw;
					} else {
						ans = ans + opposite(nw) + nw;
					}
				}
				System.out.println(ans);
			}
		}
	}

	public static boolean periodOne(String str) {
		char c = str.charAt(0);
		for (int i = 1; i < str.length(); i++) {
			if (str.charAt(i) != c) {
				return false;
			}
		}
		return true;
	}

	public static char opposite(char c){
		if (c == '1') {
			return '0';
		}
		return '1';
	}
}