description: 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.
solution: 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 as not visited (For first DFS)
      visited =[False]*(n)
      # Fill vertices in stack according to their finishing
      # times
      for i in range(n):
          if visited[i]==False:
              fillOrder(i, visited, stack,graph)
  
      # Create a reversed graph
      gr = getTranspose(graph,n)
      
      # Mark all the vertices as not visited (For second DFS)
      visited =[False]*(n)
  
      # Now process all vertices in order defined by Stack
      # print(stack)
      cnt = 0
      while stack:
          i = stack.pop()
          if visited[i]==False:
              DFSUtil(i, visited,gr)
              pp = xx[:]
              for j in xx:
                  dicti[j]=cnt
              xx = []
              cnt += 1
      newg = [[] for i in range(n)]
      for i in range(n):
          for j in gr[i]:
              newg[dicti[i]].append(dicti[j])
      for i in range(n):
          newg[i]=list(set(newg[i]))
      # print(newg)
      node = cnt-1
      # print(node)
      vis = [False]*n 
      vis[node]=True 
      st = [node]
      while st:
          node = st.pop(-1)
          for kid in newg[node]:
              if vis[kid]==False:
                  vis[kid]=True 
                  st.append(kid)
      res = ['0']*n 
      if vis.count("False")==0:
          for i in pp:
              res[i]='1'
      print(''.join(res))
      
  
  
  
  def solve():
      n = ii()
      a = li()
      b = li()
      first = []
      second = []
      first = a[:]
      second = b[:]
      d1 = {}
      d2 = {}
      e = defaultdict(lambda:0)
      for i in range(n):
          d1[a[i]]=i
          d2[b[i]]=i
  
      first.sort()
      second.sort()
      g = [[] for i in range(n)]
      # print(first)
      for i in range(n):
          p1 = bisect_right(first,a[i])
  
          if p1<n:
              p1 = d1[first[p1]]
              if e[(i,p1)]==0:
                  g[i].append(p1)
                  e[(i,p1)]=1
  
          
          p2 = bisect_right(second,b[i])
          if p2<n:
              p2 = d2[second[p2]]
              # print(i,p2)
              if e[(i,p2)]==0:
                  g[i].append(p2)
                  e[(i,p2)]=0
      
      # for i in g:
      #     print(*i)
      printSCCs(n,g)
      # print("______________")
  
  
              
  
  
  t = 1
  t = ii()
  for _ in range(t):
      solve()
  