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.
"""
    Author : thekushalghosh
    Team   : CodeDiggers
"""
import sys,math,cmath,time
start_time = time.time()

################# ---- USER DEFINED INPUT FUNCTIONS ---- #################
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(s[:len(s)])
def invr():
    return(map(int,input().split()))
##########################################################################
################# ---- THE ACTUAL CODE STARTS BELOW ---- #################

def solve():
    s = insr()
    a = list(s)
    q = a.count("1")
    w = a.count("0")
    if q == 0:
        qw = "0" * len(a)
    elif w == 0:
        qw = "1" * len(a)
    else:
        if a[0] == "0":
            qw = "01" * (q + w)
        else:
            qw = "10" * (q + w)
    print(qw)
    
################## ---- THE ACTUAL CODE ENDS ABOVE ---- ##################
##########################################################################
def main():
    ONLINE_JUDGE = __debug__
    if not ONLINE_JUDGE:
        sys.stdin = open('input.txt','r')
        sys.stdout = open('output.txt','w')
    else:
        qw = 1
    t = 1
    t = inp()
    for tt in range(t):
        solve()
    if not ONLINE_JUDGE:
        print("Time Elapsed:",time.time() - start_time,"seconds")

############## ---- USER DEFINED PROGRAMMING FUNCTIONS ---- ##############
def counter(a):
    q = [0] * max(a)
    for i in range(len(a)):
        q[a[i] - 1] = q[a[i] - 1] + 1
    return(q)

main()
sys.stdout.close()
