1342_B. Binary Period
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.
```python
import sys

t=int(sys.stdin.readline())

for _ in range(t):
    a=input()
    a=a[::]
    a=list(map(int, a))
    i=0
    if a.count(0)==0 or a.count(1)==0:
        for j in range(len(a)): 
            print(a[j], end='')
        print()
    else:
        while True:
            if i==len(a)-1:
                break
            if a[i]==0 and a[i+1]==0:
                a.insert(i+1,1)
            elif a[i]==1 and a[i+1]==1:
                a.insert(i+1,0)
            i+=1
        for k in range(len(a)): 
            print(a[k], end='')
        print()
```