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.io.*;
import java.util.*;
import java.math.BigInteger; 

public class test1 {
    
    
   static class sort implements Comparator<int[]>
   {
        public int compare(int[] a,int[] b)
        {
            //if(a[0] == b[0]) return a[1]-b[1];
           return a[0]-b[0];
        }
   }
   public static void pr(PrintWriter out,Object o)
   {
      
    out.println(o.toString());out.flush();
    
   }
   public static void prW(PrintWriter out,Object o)
   {
      
    out.print(o.toString());out.flush();
    
   }
   public static int intIn(String st)
   {
      return Integer.parseInt(st);
   }
   
   
    
    public static void pr(Object o)
    {
        System.out.println(o.toString());
    }
    public static void prW(Object o)
    {
        System.out.print(o.toString());
    }
    
   
    public static int inInt(String s)
    {
        return Integer.parseInt(s);
    }
    public static long in(String s)
    {
        return Long.parseLong(s);
    }

    
    
    static int[] toIntArray(String[] m) 
    { 
        int[] p=new int[m.length];
        for(int o=0;o<m.length;o++)
        {
            p[o]= inInt(m[o]);
        }
        return p;
    }
    static long[] toLArray(String[] m) 
    { 
       long[] p=new long[m.length];
        for(int o=0;o<m.length;o++)
        {
            p[o]= in(m[o]);
        }
        return p;
    }
    static long max=10000000000000l;static int[] before;static int R;
    public static long F(int i,int k,int l,int[] R,int[] G,int[] B,Long[][][] dp)
    {
        int n,m,t;
        n=R.length;m=G.length;t=B.length;
        i=Math.min(R.length,i);
        k=Math.min(G.length,k);
        l=Math.min(B.length,l);
        if(dp[i][k][l] != null) return dp[i][k][l];

        long max = 0l;


        if(i<n && k<m)
        {
            long va = (R[i]+0l)*(G[k]+0l);
            max=Math.max(max,va+F(i+1,k+1,l,R,G,B,dp));
        }
        if(i<n && l<t)
        {
            long va = (R[i]+0l)*(B[l]+0l);
            max=Math.max(max,va+F(i+1,k,l+1,R,G,B,dp));
        }
       if(k<m && l<t)
        {
            long va = (G[k]+0l)*(B[l]+0l);
            max=Math.max(max,va+F(i,k+1,l+1,R,G,B,dp));
        }
        dp[i][k][l]=max;return max;

        
    }
    public static String[] F(BufferedReader bf) throws Exception
    {
        return (bf.readLine().split(" "));
    }
    
    public static void main (String[] args) throws Exception {
        
        BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
       // PrintWriter out = new PrintWriter(System.out);;;
        
        //int[] map=new int[1000001];
      
        int y=1;//Integer.parseInt(bf.readLine());
        
        for(int w=0;w<y;w++)
        {

            
    
            String[] xlp = bf.readLine().split(" ");
            
            int n,k,t,h;//boolean bol=false;
           int m;
            n=inInt(xlp[0]);m=inInt(xlp[1]);k=inInt(xlp[2]);
            
             int[] R = toIntArray(F(bf));
             int[] G = toIntArray(F(bf));
             int[] B = toIntArray(F(bf));
             for(int o=0;o<n;o++) R[o]=(-1*(R[o]));
                for(int o=0;o<m;o++) G[o]=(-1*(G[o]));
                    for(int o=0;o<k;o++) B[o]=(-1*(B[o]));
             Arrays.sort(R);Arrays.sort(G);Arrays.sort(B);
            Long[][][] dp = new Long[n+1][m+1][k+1];
            for(int o=0;o<n;o++) R[o]=(-1*(R[o]));
                for(int o=0;o<m;o++) G[o]=(-1*(G[o]));
                    for(int o=0;o<k;o++) B[o]=(-1*(B[o]));
            long ans=  F(0,0,0,R,G,B,dp);
            pr(ans);
            
          
           
            

            

            
            
            
        }
        //out.close();

    }
}


/*
1
7
8 4 7 -2 20 30 -3
1 2
1 3
3 4 
4 5
4 6
3 7

1
5
8 4 7 -2 
1 2
1 3
3 4 
4 5



100663319,201326611,402653189,805306457,1610612741
Kickstart
String rp;
            rp = "Case #"+(w+1)+": "+(n-ans)+" ";
            if(curr == max) rp=rp +"INDEFINITELY";
            else rp = rp+(curr);
static int[][] dir={{0,1},{1,0},{-1,0},{0,-1}};

static class SegmentTreeRMQ 
    { 
    int st[]; 
    int minVal(int x, int y) { 
        return (x > y) ? x : y; 
    } 
  
    
    int getMid(int s, int e) { 
        return s + (e - s) / 2; 
    } 
  
    
    int RMQUtil(int ss, int se, int qs, int qe, int index) 
    { 
        
        if (qs <= ss && qe >= se) 
            return st[index]; 
  
        // If segment of this node is outside the given range 
        if (se < qs || ss > qe) 
            return Integer.MIN_VALUE; 
  
        // If a part of this segment overlaps with the given range 
        int mid = getMid(ss, se); 
        return minVal(RMQUtil(ss, mid, qs, qe, 2 * index + 1), 
                RMQUtil(mid + 1, se, qs, qe, 2 * index + 2)); 
    } 
  
    // Return minimum of elements in range from index qs (query 
    // start) to qe (query end).  It mainly uses RMQUtil() 
    int RMQ(int n, int qs, int qe) 
    { 
        // Check for erroneous input values 
        
  
        return RMQUtil(0, n - 1, qs, qe, 0); 
    } 
  
    // A recursive function that constructs Segment Tree for 
    // array[ss..se]. si is index of current node in segment tree st 
    int constructSTUtil(int arr[], int ss, int se, int si) 
    { 
        // If there is one element in array, store it in current 
        //  node of segment tree and return 
        if (ss == se) { 
            st[si] = arr[ss]; 
            return arr[ss]; 
        } 
  
        // If there are more than one elements, then recur for left and 
        // right subtrees and store the minimum of two values in this node 
        int mid = getMid(ss, se); 
        st[si] = minVal(constructSTUtil(arr, ss, mid, si * 2 + 1), 
                constructSTUtil(arr, mid + 1, se, si * 2 + 2)); 
        return st[si]; 
    } 
  
    
    void con(int arr[]) 
    { 
        // Allocate memory for segment tree 
  
        //Height of segment tree 
        int n = (arr.length);
        int x = (int) (Math.ceil(Math.log(n) / Math.log(2))); 
  
        //Maximum size of segment tree 
        int max_size = 2 * (int) Math.pow(2, x) - 1; 
        st = new int[max_size]; // allocate memory 
  
        // Fill the allocated memory st 
        constructSTUtil(arr, 0, n - 1, 0); 
    }
    }
     static class DSU {
    
    int[] p;int[] sz;int op;int c;;
    int[] last;
    public void G(int n)
    {
        last=new int[n];
        p=new int[n];
        sz=new int[n];c=n;
        op=n;
        for(int h=0;h<n;h++)
        {
            sz[h]=1;p[h]=h;
            last[h]=h;
        }
    }
    public int find(int x)
    {
        int y=x;
        while(x!=p[x]) x=p[x];
        while(y!=p[y])
        {
            int tem=p[y];
            p[y]=x;y=tem;
        }
        return p[y];
    }
    public void union(int a,int b)
    {
        int x,y;
        x=find(a);y=find(b);
        if(x==y) return;
        if(sz[x]>sz[y])
        {
            p[y] = x;
            sz[x]+=sz[y];
            last[x]=Math.max(last[x],last[y]);
        }
        else
        {
            p[x]=y;sz[y]+=sz[x];
            last[y]=Math.max(last[y],last[x]);
        }
        c--;
        
    }}

    static long pow(long x, long y, long p)
      {
        long res = 1; // Initialize result
     
        x = x % p; // Update x if it is more than or
        // equal to p
     
        if (x == 0)
          return 0l; // In case x is divisible by p;
     
        while (y > 0)
        {
     
          // If y is odd, multiply x with result
          if ((y & 1) != 0)
            res = (res * x) % p;
     
          // y must be even now
          y = y >> 1; // y = y/2
          x = (x * x) % p;
        }
        return res;
      }
      static long gcd(long a, long b)
        {
          if (b == 0)
            return a;
          return gcd(b, a % b); 
        }
        static int gcd(int a, int b,int o)
        {
          if (b == 0)
            return a;
          return gcd(b, a % b,o); 
        }
    
      
    

*/


