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: import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  import java.io.*;
  import java.util.*;
   
  public class codeforces {
  	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;
  		}
  	}
   
   
  	public static void main(String args[]) {
  		if (System.getProperty("ONLINE_JUDGE") == null) {
    			// Input is a file
    			try {
  				System.setIn(new FileInputStream("input.txt"));
  				System.setOut(new PrintStream(new FileOutputStream("output.txt")));
  			} catch (Exception e) {
  				System.err.println("Error");
  			}
  		} else {
    			// Input is System.in
  		}
  
  		FastReader sc = new FastReader();
  		// Scanner sc = new Scanner(System.in);
  		//System.out.println(java.time.LocalTime.now());
  
  		StringBuilder sb = new StringBuilder();
  		int t = sc.nextInt();
  		while(t>0) {
  			
  			long w = sc.nextInt();
  			long h = sc.nextInt();
  
  			
  			int minb = Integer.MAX_VALUE;
  			int mint = Integer.MAX_VALUE;
  			int minl = Integer.MAX_VALUE;
  			int minr = Integer.MAX_VALUE;
  
  			int maxb = Integer.MIN_VALUE;
  			int maxt = Integer.MIN_VALUE;
  			int maxl = Integer.MIN_VALUE;
  			int maxr = Integer.MIN_VALUE;
  
  
  
  			int n = sc.nextInt();
  
  			for(int i = 0; i<n; i++){
  				int k = sc.nextInt();
  				if(k > maxb)maxb = k;
  
  				if(k < minb)minb = k;
  			}
  
  			 n = sc.nextInt();
  
  			for(int i = 0; i<n; i++){
  				int k = sc.nextInt();
  				if(k > maxt)maxt = k;
  
  				if(k < mint)mint = k;
  			}
  
  			n = sc.nextInt();
  
  			for(int i = 0; i<n; i++){
  				int k = sc.nextInt();
  				if(k > maxl)maxl = k;
  
  				if(k < minl)minl = k;
  			}
  
  			n = sc.nextInt();
  
  			for(int i = 0; i<n; i++){
  				int k = sc.nextInt();
  				if(k > maxr)maxr = k;
  
  				if(k < minr)minr = k;
  			}
  
  			long Baseb = maxb - minb;
  			long Baset = maxt - mint;
  			long Basel = maxl - minl;
  			long Baser = maxr - minr;
  
  			long ans = Math.max(  Math.max(Baseb, Baset)*h,  Math.max(Basel, Baser)*w);
  
  			sb.append(ans + "\n");
  			
  
  			
  			t--;
  			}
        
  		
  		
  		System.out.println(sb);
  		
   
  		
      }
  	
  	
   
   
   
   
  //////////nCr////////////////////////////////////////
  	
   
  	
   
   
  ///////// SUM OF EACH DIGIT OF A NUMBER ///////////////
  public static long digSum(long a) {
  	long sum = 0;
  	while(a>0) {
  		sum += a%10;
  		a /= 10;
  	}
  	
  	return sum;
  }
   
  ///////// TO CHECK NUMBER IS PRIME OR NOT ///////////////
  public static boolean isPrime(int n) {
  	if(n<=1)return false;
  	if(n <= 3)return true;
  	if(n%2==0 || n%3==0)return false;
  	
  	for(int i = 5; i*i<=n; i+=6) {
  		if(n%i == 0 || n%(i+2) == 0)return false;
  	}
  	
  	return true;
  }
   
  ///////// NEXT PRIME NUMBER BIGGER THAN GIVEN NUMBER ///////////////
  public static int nextPrime(int n) {
  	while(true) {
  		n++;
  		if(isPrime(n)) break;
  			
  	}
  	
  	return n;
  }
   
  ///////// GCD  ///////////////
  public static int gcd(int a, int b) {
  	if(b == 0)return a;
  	
  	return gcd(b, a%b);
  }
   
  ///////// LCM ///////////////
  public static int lcm(int a, int b) {
  	return (a*b)/gcd(a,b);
  }
  }