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.
# cook your dish here
import math,sys
import heapq
from fractions import Fraction
from collections import Counter,defaultdict
def li(): return list(map(int,sys.stdin.readline().split()))
def ls(): return list(map(int,list(input())))
def la(): return list(input())
def ii():  return int(input())
def dic(x): return defaultdict(lambda: x) 
def isPrime(n):
    i= 2
    if n == 1:
        return False
    while i <= int(math.sqrt(n)):
        if n%i == 0:
            return False
        i = i + 1
    return True
def gcd(a,b):
    if b == 0:
        return a
    return gcd(b,a%b)
def binarySearch(arr, x): 
    l,r = 0,len(arr)-1
    while l <= r: 
        mid = l + (r - l) // 2; 
        if arr[mid] == x: 
            return True 
        elif arr[mid] < x: 
            l = mid + 1
        else: 
            r = mid - 1
    return False
t = ii()
for _ in range(t):
    a = input()
    count_1 = a.count('1')
    count_0 = len(a) - count_1
    if '0'* len(a) == a or '1'*len(a)==a:
        print(2*a)
    else:
        if a[0] == '0':
            print('01'*len(a))
        else:
            print('10'*len(a))
        
    

