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.Scanner;

public class Codeforces1342B {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int claims = in.nextInt();
        in.nextLine();
        for (int i = 0; i < claims; i++) {
            StringBuilder before = new StringBuilder(in.nextLine());
            if (before.indexOf("1") == -1 || before.indexOf("0") == -1) {
                System.out.println(before);
            } else {
                if (before.length() == 2) {
                    System.out.println(before);
                } else {
                    char c;
                    if (before.charAt(0) == '1') {
                        c = '1';
                    } else {
                        c = '0';
                    }
                    System.out.print(c);
                    for (int j = 1; j < before.length(); j++) {
                        if (before.charAt(j) == c) {
                            if (c == '1') {
                                System.out.print("0" + (c));
                            } else {
                                System.out.print("1" + c);
                            }
                        } else {
                            System.out.print(before.charAt(j));
                            c = c == '1' ? '0' : '1';
                        }
                    }
                    if (before.charAt(before.length()-1)=='0'){
                        System.out.println('1');
                    }
                    else {
                        System.out.println('0');
                    }
                }
            }
        }
    }
}