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 is_period_1_or_2(string):
    if string.find('0') == -1 or string.find('1') == -1:
        return True

    partitioned_str = []
    for i in range(0, len(string) - (len(string) % 2 != 0), 2):
        partitioned_str.append(string[i:i+2])
    # print(partitioned_str)

    p_style = partitioned_str[0]
    # print(p_style)
    for p in partitioned_str:
        if p != p_style or (p.find('0') == -1 or p.find('1') == -1):
            return False

    if len(string) % 2 != 0:
        # print(p_style[0] == string[-1])
        return p_style[0] == string[-1]
    return True


def sol_str(target_str):
    if is_period_1_or_2(target_str):
        return target_str
    
    solution_string = target_str[0]
    prev_char = target_str[0]
    for i in range(1, len(target_str)):
        if prev_char == target_str[i]:
            solution_string += '0' if prev_char == '1' else '1'    
        solution_string += target_str[i]
        
        prev_char = target_str[i]

    return solution_string


num_tests = int(input())

for _ in range(num_tests):
    target_str = input()
    print(sol_str(target_str))