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:
# t = int(input())

# for i in range(t):
#     x, y = map(int, input().split())
#     a, b = map(int, input().split())

#     cost = 0

#     cost += a * (abs(x - y))

#     z = min(x, y)
#     x = y = z

#     cost += min(x * b, a * 2*x)

#     print(cost)

t = int(input())

for i in range(t):
    x = input()

    if len(set(x)) == 1:
        print(x)
        continue

    digits = {'1': '0', '0': '1'}

    res = []
    x = x + '!'
    for i in range(len(x)-1):
        if x[i] == x[i+1]:
            res.extend([x[i], digits[x[i]]])
        else:
            res.append(x[i])

    print(''.join(res))