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.
a = int(input())
for i in range(0,a):
    b = list(input())
    c = []
    d = []
    if len(b) != 1:
        count = 1
        for j in range(0,len(b)-1):
            
            end = 0
            if b[j] != b[j+1]:
                c.append([count, b[j]])
                count = 1
                end = 1
            else:
                count += 1
                end = 2
        if end == 1 :
            c.append([1,b[len(b)-1]])
        elif end == 2:
            c.append([count,b[len(b)-1]])
        if len(c) > 1:
            for k in c:
                for ss in range(0,k[0]-1):
                    if k[1] == '1':
                        d.append('1')
                        d.append('0')
                    else :
                        d.append('0')
                        d.append('1')
                d.append(k[1])
            e = ''.join(d)
            print(e)
        elif len(c) == 1:
            print(''.join(b))
            
    else:
        print(''.join(b))
        