You are given three multisets of pairs of colored sticks: 

  * R pairs of red sticks, the first pair has length r_1, the second pair has length r_2, ..., the R-th pair has length r_R; 
  * G pairs of green sticks, the first pair has length g_1, the second pair has length g_2, ..., the G-th pair has length g_G; 
  * B pairs of blue sticks, the first pair has length b_1, the second pair has length b_2, ..., the B-th pair has length b_B; 



You are constructing rectangles from these pairs of sticks with the following process: 

  1. take a pair of sticks of one color; 
  2. take a pair of sticks of another color different from the first one; 
  3. add the area of the resulting rectangle to the total area. 



Thus, you get such rectangles that their opposite sides are the same color and their adjacent sides are not the same color.

Each pair of sticks can be used at most once, some pairs can be left unused. You are not allowed to split a pair into independent sticks.

What is the maximum area you can achieve?

Input

The first line contains three integers R, G, B (1 ≤ R, G, B ≤ 200) — the number of pairs of red sticks, the number of pairs of green sticks and the number of pairs of blue sticks.

The second line contains R integers r_1, r_2, ..., r_R (1 ≤ r_i ≤ 2000) — the lengths of sticks in each pair of red sticks.

The third line contains G integers g_1, g_2, ..., g_G (1 ≤ g_i ≤ 2000) — the lengths of sticks in each pair of green sticks.

The fourth line contains B integers b_1, b_2, ..., b_B (1 ≤ b_i ≤ 2000) — the lengths of sticks in each pair of blue sticks.

Output

Print the maximum possible total area of the constructed rectangles.

Examples

Input


1 1 1
3
5
4


Output


20


Input


2 1 3
9 5
1
2 8 5


Output


99


Input


10 1 1
11 7 20 15 19 14 2 4 13 14
8
11


Output


372

Note

In the first example you can construct one of these rectangles: red and green with sides 3 and 5, red and blue with sides 3 and 4 and green and blue with sides 5 and 4. The best area of them is 4 × 5 = 20.

In the second example the best rectangles are: red/blue 9 × 8, red/blue 5 × 5, green/blue 2 × 1. So the total area is 72 + 25 + 2 = 99.

In the third example the best rectangles are: red/green 19 × 8 and red/blue 20 × 11. The total area is 152 + 220 = 372. Note that you can't construct more rectangles because you are not allowed to have both pairs taken to be the same color.
import java.util.*;
import java.io.*;

public class D_1398 {
	
	static int a, b, c;
	static int[] ar1, ar2, ar3;
	static long[][][] memo;
	public static long dp(int i, int j, int k) {
		int z = (i == a ? 0 : 1) + (j == b ? 0 : 1) + (k == c ? 0 : 1);
		if(z <= 1)
			return 0;
		if(memo[i][j][k] != -1)
			return memo[i][j][k];
		long max = 0;
		if(i == a) {
			int x = ar2[j], y = ar3[k];
			max = Math.max(max, 1l * x * y + dp(i, j + 1, k + 1));
		} else if(j == b) {
			int x = ar1[i], y = ar3[k];
			max = Math.max(max, 1l * x * y + dp(i + 1, j, k + 1));
		} else if(k == c) {
			int x = ar1[i], y = ar2[j];
			max = Math.max(max, 1l * x * y + dp(i + 1, j + 1, k));
		} else {
			long max1 = 1l * ar1[i] * ar2[j] + dp(i + 1, j + 1, k);
			long max2 = 1l * ar2[j] * ar3[k] + dp(i, j + 1, k + 1);
			long max3 = 1l * ar1[i] * ar3[k] + dp(i + 1, j, k + 1);
			max = Math.max(max1, Math.max(max2, max3));
		}
		
		return memo[i][j][k] = max;
	}
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		
		a = sc.nextInt();
		b = sc.nextInt();
		c = sc.nextInt();
		PriorityQueue<Integer> qu1 = new PriorityQueue<>(Collections.reverseOrder());
		PriorityQueue<Integer> qu2 = new PriorityQueue<>(Collections.reverseOrder());
		PriorityQueue<Integer> qu3 = new PriorityQueue<>(Collections.reverseOrder());
		
		for(int i = 0; i < a; i++)
			qu1.add(sc.nextInt());
		for(int i = 0; i < b; i++)
			qu2.add(sc.nextInt());
		for(int i = 0; i < c; i++)
			qu3.add(sc.nextInt());
		
		ar1 = new int[a];
		ar2 = new int[b];
		ar3 = new int[c];
		for(int i = 0; i < a; i++)
			ar1[i] = qu1.poll();
		for(int i = 0; i < b; i++)
			ar2[i] = qu2.poll();
		for(int i = 0; i < c; i++)
			ar3[i] = qu3.poll();
		
		memo = new long[a + 1][b + 1][c + 1];
		for(long[][] i : memo)
			for(long[] j : i)
				Arrays.fill(j, -1);
		
		pw.println(dp(0, 0, 0));
		pw.flush();
	}

	public static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public int[] nextIntArray(int n) throws IOException {
			int[] array = new int[n];
			for (int i = 0; i < n; i++)
				array[i] = nextInt();
			return array;
		}

		public Integer[] nextIntegerArray(int n) throws IOException {
			Integer[] array = new Integer[n];
			for (int i = 0; i < n; i++)
				array[i] = new Integer(nextInt());
			return array;
		}

		public long[] nextLongArray(int n) throws IOException {
			long[] array = new long[n];
			for (int i = 0; i < n; i++)
				array[i] = nextLong();
			return array;
		}

		public double[] nextDoubleArray(int n) throws IOException {
			double[] array = new double[n];
			for (int i = 0; i < n; i++)
				array[i] = nextDouble();
			return array;
		}

		public static int[] shuffle(int[] a) {
			int n = a.length;
			Random rand = new Random();
			for (int i = 0; i < n; i++) {
				int tmpIdx = rand.nextInt(n);
				int tmp = a[i];
				a[i] = a[tmpIdx];
				a[tmpIdx] = tmp;
			}
			return a;
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}
