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:
//package codeforces;

import java.util.Scanner;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.TreeMap;
import java.util.TreeSet;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Stack;
import java.util.Set;

public class q {
	// Integer.bitCount(i); count no of 1 bit in a given number
	static Scanner scn = new Scanner(System.in);

	public static void main(String[] args) {
		/* TODO Auto-generated method stub */
		// System.out.format("%.10f", ans);char c = sc.next().charAt(0);
		// code

		int t = scn.nextInt();
		while (t-- > 0) {
			String str = scn.next();
			int o = 0, z = 0, n = str.length();
			for (int i = 0; i < n; i++) {
				if (str.charAt(i) == '0')
					z++;
				else
					o++;
			}
			if (o == 0 || z == 0)
				System.out.println(str);
			else {
				StringBuilder sb = new StringBuilder();
				for(int i=0;i<n;i++)
					sb.append("10");
			System.out.println(sb.toString());
			}
		}
	}

	public static int spk(int n) {
		if (n < 5) {
			return n;
		}
		return Math.max(4 * spk(n - 5), 3 * spk(n - 4));
	}

	public static String binary(int n) {
		StringBuilder sb = new StringBuilder();
		int rem = 0;
		if (n == 1)
			return "1";
		while (n != 1) {
			rem = n % 2;
			sb.append(rem);
			n = n / 2;
			if (n == 1)
				sb.append(n);
		}
		sb.reverse();
		return sb.toString();
	}

	public static class Pair implements Comparable {
		int x;
		int y;
		int z;

		Pair(int x, int y, int z) {
			this.x = x;
			this.y = y;
			this.z = z;
		}

		public int compareTo(Object o) {
			Pair pp = (Pair) o;
			if (pp.x == x)
				return 0;
			else if (x > pp.x)
				return 1;
			else
				return -1;
		}

	}

}