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:
#!/usr/bin/python3

# Codeforces - Educational Round 86
# Problem B - Binary Period


def read_int():
	n = int(input())
	return n


def read_ints():
	ints = [int(x) for x in input().split(" ")]
	return ints
	

#---

def solve(t):
	lmax = len(t) * 2
	sol = ''
	
	if t.find('0') == -1:
		period = '1'
		
	elif t.find('1') == -1:
		period = '0'

	elif t.find('01') == -1:
		period = '10'
	
	else:
		period = '01'
	
	lp = len(period)
	if lp == 1:
		return len(t) * period
	
	i = 0
	for j in range(len(t)):
		idx = i % 2
		val = period[idx]
		if val != t[j]:
			sol += val
			i += 1
		sol += t[j]
		i += 1
	
	return sol
			
	

# Main
T = read_int()
for case in range(T):
	t = input()
	sol = solve(t)
	
	print (sol)