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 isSubSequence(string1, string2,m,n): 
   
    # Base Cases 
    if m == 0:    return True
    if n == 0:    return False
  
    # If last characters of two strings are matching 
    if string1[m-1] == string2[n-1]: 
        return isSubSequence(string1, string2, m-1, n-1) 
  
    # If last characters are not matching 
    return isSubSequence(string1, string2, m, n-1) 

for _ in range(int(input())):
  s=input()
  x=set(s)
  if len(x)==1:
    print(s)
  else:
    l=[]
    l1=[]
    for i in range(2*len(s)):
      if i%2==0:
        l.append('1')
      else:
        l.append('0')  
    for i in range(2*len(s)):
      if i%2==0:
        l1.append('0')
      else:
        l1.append('1')
    s1=''
    s1=s1.join(l)
    s2=''
    s2=s2.join(l1)
    n=len(s)
    n1=len(s1)
    n2=len(s2)
    x=isSubSequence(s,s1,n,n1)
    if x==1:
      print(s1)
    else:
      print(s2)    
