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: //package currentContest;
  
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.ArrayDeque;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.Collections;
  import java.util.HashMap;
  import java.util.Random;
  import java.util.StringTokenizer;
  
  
  public class P1 {
  	public static void main(String[] args) {
  		// TODO Auto-generated method stub
  		FastReader sc=new FastReader();
  		int t=sc.nextInt();
  		StringBuilder st=new StringBuilder();
  		while(t--!=0){
  			long w,h;
  			w=sc.nextInt();
  			h=sc.nextInt();
  			ArrayList<ArrayList<Integer>> list=new ArrayList<ArrayList<Integer>>();
  			for(int i=0;i<4;i++) {
  				int k=sc.nextInt();
  				list.add(new ArrayList<Integer>());
  				for(int j=0;j<k;j++) {
  					list.get(i).add(sc.nextInt());
  				}
  			}
  			ArrayList<Long> ans=new ArrayList<Long>();
  			ans.add(h*(list.get(0).get(list.get(0).size()-1)-list.get(0).get(0)));
  			ans.add(h*(list.get(1).get(list.get(1).size()-1)-list.get(1).get(0)));
  			ans.add(w*(list.get(2).get(list.get(2).size()-1)-list.get(2).get(0)));
  			ans.add(w*(list.get(3).get(list.get(3).size()-1)-list.get(3).get(0)));
  			Collections.sort(ans);
  			st.append(ans.get(3)+"\n");
  			
  		}
  		System.out.println(st);
  	}
  	
  	static class DisjointUnionSets {
  		int[] rank, parent;
  		int n;
  		
  		// Constructor
  		public DisjointUnionSets(int n)
  		{
  			rank = new int[n];
  			parent = new int[n];
  			this.n = n;
  			makeSet();
  		}
  		
  		// Creates n sets with single item in each
  		void makeSet()
  		{
  			for (int i = 0; i < n; i++) {
  				// Initially, all elements are in
  				// their own set.
  				parent[i] = i;
  			}
  		}
  		
  		// Returns representative of x's set
  		int find(int x)
  		{
  			// Finds the representative of the set
  			// that x is an element of
  			if (parent[x] != x) {
  				// if x is not the parent of itself
  				// Then x is not the representative of
  				// his set,
  				parent[x] = find(parent[x]);
  				
  				// so we recursively call Find on its parent
  				// and move i's node directly under the
  				// representative of this set
  			}
  			
  			return parent[x];
  		}
  		
  		// Unites the set that includes x and the set
  		// that includes x
  		void union(int x, int y)
  		{
  			// Find representatives of two sets
  			int xRoot = find(x), yRoot = find(y);
  			
  			// Elements are in the same set, no need
  			// to unite anything.
  			if (xRoot == yRoot)
  				return;
  			
  			// If x's rank is less than y's rank
  			if (rank[xRoot] < rank[yRoot])
  				
  				// Then move x under y  so that depth
  				// of tree remains less
  				parent[xRoot] = yRoot;
  			
  			// Else if y's rank is less than x's rank
  			else if (rank[yRoot] < rank[xRoot])
  				
  				// Then move y under x so that depth of
  				// tree remains less
  				parent[yRoot] = xRoot;
  			
  			else // if ranks are the same
  			{
  				// Then move y under x (doesn't matter
  				// which one goes where)
  				parent[yRoot] = xRoot;
  				
  				// And increment the result tree's
  				// rank by 1
  				rank[xRoot] = rank[xRoot] + 1;
  			}
  		}
  	}
  
  	static FastReader sc = new FastReader();
  
  	public static void solvegraph() {
  		int n = sc.nextInt();
  
  		int edge[][] = new int[n - 1][2];
  		for (int i = 0; i < n - 1; i++) {
  			edge[i][0] = sc.nextInt() - 1;
  			edge[i][1] = sc.nextInt() - 1;
  		}
  		ArrayList<ArrayList<Integer>> ad = new ArrayList<>();
  		for (int i = 0; i < n; i++) {
  			ad.add(new ArrayList<Integer>());
  		}
  		for (int i = 0; i < n - 1; i++) {
  			ad.get(edge[i][0]).add(edge[i][1]);
  			ad.get(edge[i][1]).add(edge[i][0]);
  		}
  		int parent[] = new int[n];
  		Arrays.fill(parent, -1);
  		parent[0] = n;
  		ArrayDeque<Integer> queue = new ArrayDeque<>();
  		queue.add(0);
  		int child[] = new int[n];
  		Arrays.fill(child, 0);
  		ArrayList<Integer> lv = new ArrayList<Integer>();
  		while (!queue.isEmpty()) {
  			int toget = queue.getFirst();
  			queue.removeFirst();
  			child[toget] = ad.get(toget).size() - 1;
  			for (int i = 0; i < ad.get(toget).size(); i++) {
  				if (parent[ad.get(toget).get(i)] == -1) {
  					parent[ad.get(toget).get(i)] = toget;
  					queue.addLast(ad.get(toget).get(i));
  				}
  			}
  			lv.add(toget);
  		}
  
  		child[0]++;
  	}
  
  	static void sort(int[] A) {
  		int n = A.length;
  		Random rnd = new Random();
  		for (int i = 0; i < n; ++i) {
  			int tmp = A[i];
  			int randomPos = i + rnd.nextInt(n - i);
  			A[i] = A[randomPos];
  			A[randomPos] = tmp;
  		}
  		Arrays.sort(A);
  	}
  	static void sort(char[] A) {
  		int n = A.length;
  		Random rnd = new Random();
  		for (int i = 0; i < n; ++i) {
  			char tmp = A[i];
  			int randomPos = i + rnd.nextInt(n - i);
  			A[i] = A[randomPos];
  			A[randomPos] = tmp;
  		}
  		Arrays.sort(A);
  	}
  
  	static void sort(long[] A) {
  		int n = A.length;
  		Random rnd = new Random();
  		for (int i = 0; i < n; ++i) {
  			long tmp = A[i];
  			int randomPos = i + rnd.nextInt(n - i);
  			A[i] = A[randomPos];
  			A[randomPos] = tmp;
  		}
  		Arrays.sort(A);
  	}
  
  	static String sort(String s) {
  		Character ch[] = new Character[s.length()];
  		for (int i = 0; i < s.length(); i++) {
  			ch[i] = s.charAt(i);
  		}
  		Arrays.sort(ch);
  		StringBuffer st = new StringBuffer("");
  		for (int i = 0; i < s.length(); i++) {
  			st.append(ch[i]);
  		}
  		return st.toString();
  	}
  
  	public static long gcd(long a, long b) {
  		if (a == 0) {
  			return b;
  		}
  		return gcd(b % a, a);
  	}
  
  	static class FastReader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastReader() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
  
  		String next() {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
  
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  
  		String nextLine() {
  			String str = "";
  			try {
  				str = br.readLine();
  			} catch (IOException e) {
  				e.printStackTrace();
  			}
  			return str;
  		}
  	}
  }
  