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.*;
import java.io.*;
import java.math.BigInteger;
public class bver{
public static long mod = 1000000007;
public static int dx[] = {-1,0,1,0,-1,-1,1,1,-1};//primero ady, despues esq
public static int dy[] = {0,1,0,-1,1,1,-1,-1};
public static void main(String args[]) throws IOException{
StringBuilder sb = new StringBuilder("");
BufferedReader lector = new BufferedReader(new InputStreamReader(System.in));
//String tmp[] = lector.readLine().split(" ");
int cas = Integer.parseInt(lector.readLine());
for(int n = 0;n<cas;n++){
String t = lector.readLine();
if(t.length()<=2){
sb.append(t+"\n");
continue;
}
String tt = "";
int unos = 0;
for(int k = 0;k<t.length();k++){
String ss = "";
if(t.charAt(k)=='1'){
unos++;
ss = "10";
}else
ss = "10";
tt +=ss;
}
sb.append(((unos==0 ||unos==t.length())?t:tt)+"\n");
}
System.out.print(sb);
}
public static int atoi(String a){
return Integer.parseInt(a);
}
public static long atol(String a){
return Long.parseLong(a);
}
}
class p implements Comparable{
public int a;
public p(int a){
this.a = a;
}
public String toString(){
return "";
}
public int compareTo(Object o){
p pp = (p)o;
return (int)Math.signum(a-pp.a);
}
}
//sergio1
