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.*;
import java.math.BigInteger;
public class cf1 {
    static long mod = (long)1e9 + 7;
    static long mod1 = 998244353;
    static FastScanner f;
    static PrintWriter pw = new PrintWriter(System.out);
    static Scanner S = new Scanner(System.in);
    static long x0; static long y0;
    static int inf = (int)(1e9) + 5;
    static long iinf = (long)(1e18);
    static void solve()throws IOException {
       int r = f.ni(); int g = f.ni(); int b = f.ni();
       long red[] = new long[r]; long green[] = new long[g]; long blue[] = new long[b];
       for (int i = 0; i < r; ++i) red[i] = f.nl(); sort(red);
       for (int i = 0; i < g; ++i) green[i] = f.nl(); sort(green);
       for (int i = 0; i < b; ++i) blue[i] = f.nl(); sort(blue);

       long dp[][][] = new long[r + 1][g + 1][b + 1];
       for (int i = 0; i <= r; ++i) {
            for (int j = 0; j <= g; ++j) {
                for (int k = 0; k <= b; ++k) {
                    if (i + j + k <= 1) continue;
                    long c1 = 0 , c2 = 0 , c3 = 0;
                    if (i > 0 && j > 0)
                        c1 = dp[i - 1][j - 1][k] + red[i - 1] * green[j - 1];
                    if (i > 0 && k > 0) 
                        c2 = dp[i - 1][j][k - 1] + red[i - 1] * blue[k - 1];
                    if (j > 0 && k > 0)
                        c3 = dp[i][j - 1][k - 1] + green[j - 1] * blue[k - 1];
                    if (i > 0) dp[i][j][k] = Math.max(dp[i][j][k] , dp[i - 1][j][k]);
                    if (j > 0) dp[i][j][k] = Math.max(dp[i][j][k] , dp[i][j - 1][k]);
                    if (k > 0) dp[i][j][k] = Math.max(dp[i][j][k] , dp[i][j][k - 1]);

                    dp[i][j][k] = Math.max(dp[i][j][k] , c1);
                    dp[i][j][k] = Math.max(dp[i][j][k] , c2);
                    dp[i][j][k] = Math.max(dp[i][j][k] , c3);
                }
            }
       }
       pn(dp[r][g][b]);
    }   
    public static void main(String[] args)throws NumberFormatException , IOException {
        init();
        boolean tc = false;
        int t = tc ? f.ni() : 1;
        while(t --> 0) solve();
        pw.flush(); 
        pw.close();  
    }
        
/******************************END OF MAIN PROGRAM*******************************************/
    public static void init()throws IOException{if(System.getProperty("ONLINE_JUDGE")==null){f=new FastScanner("");}else{f=new FastScanner(System.in);}}
    public static class FastScanner {
        BufferedReader br;StringTokenizer st;
        FastScanner(InputStream stream){try{br=new BufferedReader(new InputStreamReader(stream));}catch(Exception e){e.printStackTrace();}}
        FastScanner(String str){try{br=new BufferedReader(new FileReader("!a.txt"));}catch(Exception e){e.printStackTrace();}}
        String next(){while(st==null||!st.hasMoreTokens()){try{st=new StringTokenizer(br.readLine());}catch(IOException e){e.printStackTrace();}}return st.nextToken();}
        String nextLine()throws IOException{return br.readLine();}int ni(){return Integer.parseInt(next());}long nl(){return Long.parseLong(next());}double nd(){return Double.parseDouble(next());}
    }
    public static void pn(Object o){pw.println(o);}
    public static void p(Object o){pw.print(o);}
    public static void pni(Object o){pw.println(o);pw.flush();}
    static int gcd(int a,int b){if(b==0)return a;else{return gcd(b,a%b);}}
    static long gcd(long a,long b){if(b==0l)return a;else{return gcd(b,a%b);}}
    static long lcm(long a,long b){return (a*b/gcd(a,b));}
    static long exgcd(long a,long b){if(b==0){x0=1;y0=0;return a;}long temp=exgcd(b,a%b);long t=x0;x0=y0;y0=t-a/b*y0;return temp;}
    static long pow(long a,long b){long res=1;while(b>0){if((b&1)==1)res=res*a;b>>=1;a=a*a;}return res;}
    static long mpow(long a,long b){long res=1;while(b>0l){if((b&1)==1l)res=((res%mod)*(a%mod))%mod;b>>=1l;a=((a%mod)*(a%mod))%mod;}return res;}
    static long mul(long a , long b){return ((a%mod)*(b%mod)%mod);}
    static long adp(long a , long b){return ((a%mod)+(b%mod)%mod);}
    static int log2(int x){return (int)(Math.log(x)/Math.log(2));}
    static boolean isPrime(long n){if(n<=1)return false;if(n<=3)return true;if(n%2==0||n%3==0)return false;for(long i=5;i*i<=n;i=i+6)if(n%i==0||n%(i+2)==0)return false;return true;}
    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=i+6)if(n%i==0||n%(i+2)==0)return false;return true;}
    static HashSet<Long> factors(long n){HashSet<Long> hs=new HashSet<Long>();for(long i=1;i<=(long)Math.sqrt(n);i++){if(n%i==0){hs.add(i);hs.add(n/i);}}return hs;}
    static HashSet<Integer> factors(int n){HashSet<Integer> hs=new HashSet<Integer>();for(int i=1;i<=(int)Math.sqrt(n);i++){if(n%i==0){hs.add(i);hs.add(n/i);}}return hs;}
    static HashSet<Long> pf(long n){HashSet<Long> ff=factors(n);HashSet<Long> res=new HashSet<Long>();for(Long i:ff)if(isPrime(i))res.add(i);return res;}
    static HashSet<Integer> pf(int n){HashSet<Integer> ff=factors(n);HashSet<Integer> res=new HashSet<Integer>();for(Integer i:ff)if(isPrime(i))res.add(i);return res;}
    static int[] inpint(int n){int arr[]=new int[n+1];for(int i=1;i<=n;++i){arr[i]=f.ni();}return arr;}
    static long[] inplong(int n){long arr[] = new long[n+1];for(int i=1;i<=n;++i){arr[i]=f.nl();}return arr;}
    static boolean ise(int x){return ((x&1)==0);}static boolean ise(long x){return ((x&1)==0);}
    static int gnv(char c){return Character.getNumericValue(c);}//No. of integers less than equal to i in ub
    static int log(long x){return x==1?0:(1+log(x/2));} static int log(int x){return x==1?0:(1+log(x/2));}
    static int upperbound(int a[],int i){int lo=0,hi=a.length-1,mid=0;int count=0;while(lo<=hi){mid=(lo+hi)/2;if(a[mid]<=i){count=mid+1;lo=mid+1;}else hi=mid-1;}return count;}
    static void sort(int[] a){ArrayList<Integer> l=new ArrayList<>();for(int i:a)l.add(i);Collections.sort(l);for(int i=0;i<a.length;++i)a[i]=l.get(i);}
    static void sort(long[] a){ArrayList<Long> l=new ArrayList<>();for(long i:a)l.add(i);Collections.sort(l);for(int i=0;i<a.length;++i)a[i]=l.get(i);}
    static void sort(ArrayList<Integer> a){Collections.sort(a);}//!Precompute fact in ncr()!
    static int nextPowerOf2(int n){int count=0;if(n>0&&(n&(n-1))==0)return n;while(n!=0){n>>=1;count += 1;}return 1<<count;} 
}