human
Write me a Python code to determine if he can win the tournament given the following:
Can you solve the below in PYTHON3?
n players are playing a game. 

There are two different maps in the game. For each player, we know his strength on each map. When two players fight on a specific map, the player with higher strength on that map always wins. No two players have the same strength on the same map. 

You are the game master and want to organize a tournament. There will be a total of n-1 battles. While there is more than one player in the tournament, choose any map and any two remaining players to fight on it. The player who loses will be eliminated from the tournament. 

In the end, exactly one player will remain, and he is declared the winner of the tournament. For each player determine if he can win the tournament.

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of players.

The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9, a_i ≠ a_j for i ≠ j), where a_i is the strength of the i-th player on the first map. 

The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9, b_i ≠ b_j for i ≠ j), where b_i is the strength of the i-th player on the second map. 

It is guaranteed that the sum of n over all test cases does not exceed 10^5.

Output

For each test case print a string of length n. i-th character should be "1" if the i-th player can win the tournament, or "0" otherwise.

Example

Input

3
4
1 2 3 4
1 2 3 4
4
11 12 20 21
44 22 11 30
1
1000000000
1000000000

Output

0001
1111
1

Note

In the first test case, the 4-th player will beat any other player on any game, so he will definitely win the tournament.

In the second test case, everyone can be a winner. 

In the third test case, there is only one player. Clearly, he will win the tournament.
assistant
#Here is Python code to determine if he can win the tournament.

import sys,os,io
import math 
from bisect import bisect_right
from collections import defaultdict
def ii():
    return int(input())
def li():
    return list(map(int,input().split()))
if(os.path.exists('input.txt')):
    sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w") 
else:
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

#for deep recursion__________________________________________-
from types import GeneratorType
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to

    return wrappedfunc

xx = []
pp = []
dicti = defaultdict(lambda:-1)
cnt = 0
# A function used by DFS

@bootstrap
def DFSUtil(v,visited,graph):
    global xx
    # Mark the current node as visited and print it
    visited[v]= True
    xx.append(v)
    #Recur for all the vertices adjacent to this vertex
    for i in graph[v]:
        if visited[i]==False:
            yield DFSUtil(i,visited,graph)
    yield

@bootstrap
def fillOrder(v,visited, stack,graph):
    visited[v]= True
    for i in graph[v]:
        if visited[i]==False:
            yield fillOrder(i, visited, stack,graph)
    stack = stack.append(v)
    yield
    

# Function that returns reverse (or transpose) of this graph
def getTranspose(graph,n):
    tg = [[] for i in range(n)]
    # Recur for all the vertices adjacent to this vertex
    for i in range(len(graph)):
        for j in graph[i]:
            tg[j].append(i)
    return tg


# The main function that finds and prints all strongly
# connected components
def printSCCs(n,graph):
    global xx,pp,cnt
    stack = []
    # Mark all the vertices...