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 PYTHON3):

Solution:
import math
num_case = int(input())
def invert(char):
	return '1' if char=='0' else '0'
for i in range(num_case):
	se = input()
	if se=="00" or se=="01" or se=="10" or se=="11":
		print(se)
	else:
		new_str = ''
		temp = True
		for i in range(len(se)-1):
			if se[i]!=se[i+1]:
				temp=False
		if temp==True:
			print(se+se)
		else:
			for i in range(len(se)-1):
				new_str+=se[i]
				if se[i]==se[i+1]:
					new_str+=invert(se[i])
			c = len(se)-1
			new_str+=se[c]
			if len(new_str)%2==1:
				new_str+=invert(se[c])
			print(new_str)