Problem: 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.

Solution (in PYTHON3):

Solution:
from collections import defaultdict as dc
from heapq import *
import math
import bisect
import sys
from collections import deque as dq
from heapq import heapify,heappush,heappop
mod=10**9 +7
def inp(): 
    p=int(input())
    return p
def line():
    p=list(map(int,input().split()))
    return p
def read_mat():
    n=inp()
    a=[]
    for i in range(n):
        a.append(line())
    return a
def digit(n):
    s=str(n)
    p=0
    for i in s:
        p+=(int(i))**2
    return p

def solve(s):
    n=len(s)
    z=s.count('0')
    if z==n or z==0:
        return s
    p=''
    for i in range(n-1):
        if s[i]=='0' and s[i+1]=='0':
            p+='01'
        elif s[i]=='1' and s[i+1]=='1':
            p+='10'
        else:
            p+=s[i]
    p=p+s[n-1]
    return p

for _ in range(inp()):
    s=str(input())
    print(solve(s))