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.


# def find(x, y, a, b):
# 	if x == 0:
# 		return a * abs(y)
# 	elif y == 0:
# 		return a * abs(x)
# 	elif x * y < 0:
# 		return abs(x) * a + abs(y) * a
# 	# знак один
# 	else:
# 		diff = abs(x - y)
# 		sec_oper_it = min(abs(x), abs(y))
# 		second_operation_cost = sec_oper_it * b
# 		first_operation_cost = diff * a 
# 		only_first_operation_cost = abs(x) * a + abs(y) * a
# 		return min(only_first_operation_cost, second_operation_cost + first_operation_cost)




# cases = int(input())
# for _ in range(cases):
# 	x, y = list(map(int, input().split()))
# 	a, b = list(map(int, input().split()))
# 	print(find(x, y, a, b))
	
def solve(t):
    if len(t) == 2:
        return t
    if len(set(t)) == 1:
        return t
    s = [t[0]]
    for cur in t[1:]:
        if cur == s[-1]:
            s.append((cur + 1) % 2)
        s.append(cur)
    return s


cases = int(input())
for _ in range(cases):
	t = list(map(int, list(input())))
	print(''.join(str(k) for k in solve(t)))
