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:
res = []
def rev(s):
    if s == "1":
        return "0"
    return "1"
for _ in range(int(input())):
    t = input()
    l = len(t)
    if len(set(t)) == 1:
        res.append(t)
    else:
        j = 0
        r = []
        for i in range(2*l):
            if i % 2 == 0:
                r.append(t[j])
                j += 1
            else:
                r.append("")
                
        for i in range(l):
            if i == l-1:
                if (2*l - r.count("")) % 2:
                    r[2*i+1] = rev(t[i])
            elif t[i] == t[i+1]:
                r[2*i+1] = rev(t[i])
        res.append("".join(r))
                
for i in res:
    print(i)