description: A rectangle with its opposite corners in (0, 0) and (w, h) and sides parallel to the axes is drawn on a plane.
  
  You are given a list of lattice points such that each point lies on a side of a rectangle but not in its corner. Also, there are at least two points on every side of a rectangle.
  
  Your task is to choose three points in such a way that: 
  
    * exactly two of them belong to the same side of a rectangle; 
    * the area of a triangle formed by them is maximum possible. 
  
  
  
  Print the doubled area of this triangle. It can be shown that the doubled area of any triangle formed by lattice points is always an integer.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of testcases.
  
  The first line of each testcase contains two integers w and h (3 ≤ w, h ≤ 10^6) — the coordinates of the corner of a rectangle.
  
  The next two lines contain the description of the points on two horizontal sides. First, an integer k (2 ≤ k ≤ 2 ⋅ 10^5) — the number of points. Then, k integers x_1 < x_2 < ... < x_k (0 < x_i < w) — the x coordinates of the points in the ascending order. The y coordinate for the first line is 0 and for the second line is h.
  
  The next two lines contain the description of the points on two vertical sides. First, an integer k (2 ≤ k ≤ 2 ⋅ 10^5) — the number of points. Then, k integers y_1 < y_2 < ... < y_k (0 < y_i < h) — the y coordinates of the points in the ascending order. The x coordinate for the first line is 0 and for the second line is w.
  
  The total number of points on all sides in all testcases doesn't exceed 2 ⋅ 10^5.
  
  Output
  
  For each testcase print a single integer — the doubled maximum area of a triangle formed by such three points that exactly two of them belong to the same side.
  
  Example
  
  Input
  
  
  3
  5 8
  2 1 2
  3 2 3 4
  3 1 4 6
  2 4 5
  10 7
  2 3 9
  2 1 7
  3 1 3 4
  3 4 5 6
  11 5
  3 1 6 8
  3 3 6 8
  3 1 3 4
  2 2 4
  
  
  Output
  
  
  25
  42
  35
  
  Note
  
  The points in the first testcase of the example: 
  
    * (1, 0), (2, 0); 
    * (2, 8), (3, 8), (4, 8); 
    * (0, 1), (0, 4), (0, 6); 
    * (5, 4), (5, 5). 
  
  
  
  The largest triangle is formed by points (0, 1), (0, 6) and (5, 4) — its area is 25/2. Thus, the doubled area is 25. Two points that are on the same side are: (0, 1) and (0, 6).
solution: from __future__ import division, print_function
  
  import os,sys
  from io import BytesIO, IOBase
  
  if sys.version_info[0] < 3:
      from __builtin__ import xrange as range
      from future_builtins import ascii, filter, hex, map, oct, zip
  
  from bisect import bisect_left as lower_bound, bisect_right as upper_bound 
  def so():  return int(input())
  def st():  return input()
  def mj():  return map(int, input().split())
  def msj(): return map(str,input().strip().split(" "))
  def le():  return list(map(int,input().split()))
  def rc():  return map(float,input().split())
  def lebe():return list(map(int, input()))
  
  def dmain():
      sys.setrecursionlimit(1000000)
      threading.stack_size(1024000)
      thread = threading.Thread(target=main)
      thread.start()
  def joro(L):
      return(''.join(map(str, L)))
  
  
  def decimalToBinary(n): return bin(n).replace("0b","")
  
  
  def isprime(n):
      for i in range(2,int(n**0.5)+1):
          if n%i==0:
              return False
      return True
  def npr(n, r):
      return factorial(n) // factorial(n - r) if n >= r else 0
   
   
  def ncr(n, r):
      return factorial(n) // (factorial(r) * factorial(n - r)) if n >= r else 0
   
   
  def lower_bound(li, num):
      answer = -1
      start = 0
      end = len(li) - 1
   
      while (start <= end):
          middle = (end + start) // 2
          if li[middle] >= num:
              answer = middle
              end = middle - 1
          else:
              start = middle + 1
      return answer  # min index where x is not less than num
   
   
  def upper_bound(li, num):
      answer = -1
      start = 0
      end = len(li) - 1
   
      while (start <= end):
          middle = (end + start) // 2
   
          if li[middle] <= num:
              answer = middle
              start = middle + 1
   
          else:
              end = middle - 1
      return answer  # max index where x is not greater than num
  def tir(a,b,c):
      if(0==c):
          return 1
      if(len(a)<=b):
          return 0
      
      if(c!=-1):
          return (tir(a,1+b,c+a[b]) or tir(a,b+1,c-a[b]) or tir(a,1+b,c)) 
          
          
      else:
          return (tir(a,1+b,a[b]) or tir(a,b+1,-a[b]) or tir(a,1+b,-1))
  hoi=int(2**20)        
   
  def abs(x):
      return x if x >= 0 else -x
   
   
  def binary_search(li, val, lb, ub):
      # print(lb, ub, li)
      ans = -1
      while (lb <= ub):
          mid = (lb + ub) // 2
          # print('mid is',mid, li[mid])
          if li[mid] > val:
              ub = mid - 1
          elif val > li[mid]:
              lb = mid + 1
          else:
              ans = mid  # return index
              break
      return ans
   
   
  def kadane(x):  # maximum sum contiguous subarray
      sum_so_far = 0
      current_sum = 0
      for i in x:
          current_sum += i
          if current_sum < 0:
              current_sum = 0
          else:
              sum_so_far = max(sum_so_far, current_sum)
      return sum_so_far
   
   
  def pref(li):
      pref_sum = [0]
      for i in li:
          pref_sum.append(pref_sum[-1] + i)
      return pref_sum
  def gcd(x, y):
      while(y):
         x, y = y, x % y
      return x
  
    
     
      
      
   
  def SieveOfEratosthenes(n):
      prime = [True for i in range(n + 1)]
      p = 2
      li = []
      while (p * p <= n):
          if (prime[p] == True):
              for i in range(p * p, n + 1, p):
                  prime[i] = False
          p += 1
   
      for p in range(2, len(prime)):
          if prime[p]:
              li.append(p)
      return li
   
   
  def primefactors(n):
      factors = []
      while (n % 2 == 0):
          factors.append(2)
          n //= 2
      for i in range(3, int(sqrt(n)) + 1, 2):  # only odd factors left
          while n % i == 0:
              factors.append(i)
              n //= i
      if n > 2:  # incase of prime
          factors.append(n)
      return factors
  
  def read():
      sys.stdin  = open('input.txt', 'r')  
      sys.stdout = open('output.txt', 'w') 
  def tr(n):
      return n*(n+1)//2
  poi=998244353
  soi=int(1e6)
  boi=int(1e9-7)
  doi=int(1e9+7)
  
  def iu():
      import sys
      import math as my
      input=sys.stdin.readline
      y="YES"
      n="NO"
      w,h=mj()
      L=le()
      M=[]
      for i in range(1,len(L)):
          M.append(L[i])
      k=L[0]
      Z=le()
      P=[]
      for i in range(1,len(Z)):
          P.append(Z[i])
      Y=le()
      Q=[]
      for i in range(1,len(Y)):
          Q.append(Y[i])
      R=[]
      X=le()
      for i in range(1,len(X)):
          R.append(X[i])
      M.sort()
      P.sort()
      Q.sort()
      R.sort()
      a=h*max(M[-1]-M[0],P[-1]-P[0])
      a=max(a,w*max(Q[-1]-Q[0],R[-1]-R[0]))
      print(a)
      
      
          
      
      
      
      
              
      
          
          
              
  def main():
      for i in range(so()):
          iu()    
      
      
      
      
  
      
  
      
              
              
              
                  
              
          
      
      
      
          
      
      
      
      
      
          
          
                          
              
      
          
      
              
      
              
          
      
  
              
          
      
      
      
  
      
      
      
      
          
  
          
      
  
      
      
      
      
      
      
      
      
      
      
  
      
  
      
          
          
              
      
      
                  
              
                  
                  
                  
                  
              
      
              
              
      
      
      
      
          
              
      
             
              
              
          
      
          
      
      
          
          
          
  
              
                      
                  
                      
                      
                      
                      
                      
                      
          
                  
          
          
         
      
             
            
            
                  
              
          
                  
      
          
          
          
         
              
                  
          
  
  
  
  
  # region fastio
  # template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
  
  BUFSIZE = 8192
  
  
  class FastIO(IOBase):
      newlines = 0
  
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
  
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
  
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
  
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  
  def print(*args, **kwargs):
      """Prints the values to a stream, or to sys.stdout by default."""
      sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
      at_start = True
      for x in args:
          if not at_start:
              file.write(sep)
          file.write(str(x))
          at_start = False
      file.write(kwargs.pop("end", "\n"))
      if kwargs.pop("flush", False):
          file.flush()
  
  
  if sys.version_info[0] < 3:
      sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
  else:
      sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  # endregion
  
  
  if __name__ == "__main__":
      #read()
      main()
      #dmain()
  
  # Comment Read()