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: # cook your dish here
  import sys
  import bisect
  from bisect import bisect_left as lb
  from bisect import bisect_right as rb
  input_=lambda: sys.stdin.readline().strip("\r\n")
  from math import log
  from math import gcd
  from math import atan2,acos
  from random import randint
  sa=lambda :input_()
  sb=lambda:int(input_())
  sc=lambda:input_().split()
  sd=lambda:list(map(int,input_().split()))
  sflo=lambda:list(map(float,input_().split()))
  se=lambda:float(input_())
  sf=lambda:list(input_())
  flsh=lambda: sys.stdout.flush()
  sys.setrecursionlimit(3*10**5)
  mod=10**9+7
  mod1=998244353
  gp=[]
  cost=[]
  dp=[]
  mx=[]
  ans1=[]
  ans2=[]
  special=[]
  specnode=[]
  a=0
  kthpar=[]
  def dfs2(root,par):
      if par!=-1:
          dp[root]=dp[par]+1
      for i in range(1,20):
          if kthpar[root][i-1]!=-1:
              kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
      for child in gp[root]:
          if child==par:continue
          kthpar[child][0]=root
          dfs(child,root)
          
  ans=0
  a=[]
  n,k=0,0
  b=[]
  vis=[]
  tot=0
  time=[]
  time1=[]
  adj=[]
  mx=-1
  eps=0.0000001
  x=0
  pref=[]
  lvl=[]
  gp=[]
  ans=[]
  f=0
  w=[]
  
  def dfs(i,prod,rem,n):
      if rem==0:
          while(i<len(n)):
              prod*=int(n[i])
              i+=1
          return prod
      if i==len(n):
          return prod
      mx=0
      for j in range(9-int(n[i])):
          mx=max(dfs(i+1,(int(n[i])+min(j,rem))*prod,rem-min(j,rem),n),mx)
      return mx
  class Disjoint_SU:
      def __init__(self,n):
          self.parent=[i for i in range(n)]
          self.size=[1 for i in range(n)]
      #find
      def lake_de(self,a):
          if a==self.parent[a]:
              return a
          self.parent[a]=self.lake_de(self.parent[a])
          return self.parent[a]
      #union
      #by size
      def jod(self,a,b):
          x=self.lake_de(a)
          y=self.lake_de(b)
          #for cycle in graph
          if x==y:
              return True
          if self.size[x]>self.size[y]:
              self.size[x]+=self.size[y]
              self.parent[y]=x
          else:
              self.size[y]+=self.size[x]
              self.parent[x]=y
          return False
  ans=0
  h,w=0,0
  def func(x1,y1,x2,y2,first,second,third,fourth):
      global ans,h,w
      for i in range(1,len(first)):
          x3,y3=first[i],0
          ans=max(ans,x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2))
      for i in range(1,len(second)):
          x3,y3=second[i],h
          ans=max(ans,x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2))
      for i in range(1,len(third)):
          x3,y3=0,third[i]
          ans=max(ans,x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2))
      for i in range(1,len(fourth)):
          x3,y3=w,fourth[i]
          ans=max(ans,x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2))
      return
  def hnbhai(tc):
      global ans,h,w
      ans=0
      x,y=0,0
      w,h=sd()
      first=sd()
      second=sd()
      third=sd()
      fourth=sd()
      ans=first[-1]-first[1]
      ans*=h
      ans=max(ans,h*(second[-1]-second[1]))
      ans=max(ans,w*(third[-1]-third[1]),w*(fourth[-1]-fourth[1]))
      print(ans)
          
      return
  for _ in range(sb()):
      hnbhai(_+1)
  ##case
  
  #.
  