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 can(s,t):
    if len(s)>2*len(t):
        return False
    ct=len(t)
    pos=0
    for i in range(len(s)):
        if s[i]==t[pos]:
            ct-=1 
            pos+=1 
        if ct==0:
            break
    return ct==0
        
for _ in range(int(input())):
    #n=int(input())
    #x,y=map(int,input().split())
    #a=list(map(int,input().split()))
    t=input()
    s=''
    ans=t 
    ansp=len(t)
    for p in range(len(t)-1,0,-1):
        for i in range(len(t)):
            curr=t[i:i+p+1]
            cs=''
            while len(cs)<2*len(t):
                cs+=curr 
            if can(cs,t) and ansp>p:
                ansp=p
                ans=cs 
    print(ans)
        
        
    
        
    