Vasya follows a basketball game and marks the distances from which each team makes a throw. He knows that each successful throw has value of either 2 or 3 points. A throw is worth 2 points if the distance it was made from doesn't exceed some value of d meters, and a throw is worth 3 points if the distance is larger than d meters, where d is some non-negative integer.

Vasya would like the advantage of the points scored by the first team (the points of the first team minus the points of the second team) to be maximum. For that he can mentally choose the value of d. Help him to do that.

Input

The first line contains integer n (1 ≤ n ≤ 2·105) — the number of throws of the first team. Then follow n integer numbers — the distances of throws ai (1 ≤ ai ≤ 2·109). 

Then follows number m (1 ≤ m ≤ 2·105) — the number of the throws of the second team. Then follow m integer numbers — the distances of throws of bi (1 ≤ bi ≤ 2·109).

Output

Print two numbers in the format a:b — the score that is possible considering the problem conditions where the result of subtraction a - b is maximum. If there are several such scores, find the one in which number a is maximum.

Examples

Input

3
1 2 3
2
5 6


Output

9:6


Input

5
6 7 8 9 10
5
1 2 3 4 5


Output

15:10




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





public class Main {
    
    
    
    
   static class sort implements Comparator<int[]>
   {
        public int compare(int[] a,int[] b)
        {
            
            return -b[0]+a[0];
        }
   }
   static class sort1 implements Comparator<long[]>
   {
        public int compare(long[] a,long[] b)
        {
            
          long c = -(a[0]-a[1]) + (b[0]-b[1]);

          if(c==0l) return 0;
          if(c>0l) return 1;
          return -1;
        }
   }
   static class sort5 implements Comparator<long[]>
   {
        public int compare(long[] a,long[] b)
        {
            
           double d1,d2;
           d1 = (a[0]+0.00)/(a[1]+0.00);
           d2 = (b[0]+0.00)/(b[1]+0.00);
           //int v = Double.comapre(d2,d1);
           double eph = 0.000000000000001;
           double d = (d2-d1);
          // pr(d1+" "+d2);
           if(d>eph) return 1;
           else if(d<eph) return -1;
           if(a[1]>b[1]) return 1;
           else if(b[1]>a[1]) return -1;
           return 0;
        }
   }
   static class sort11 implements Comparator<double[]>
   {
        public int compare(double[] a,double[] b)
        {
            //if(a[0] == b[0]) return a[1]-b[1];
           if(a[1] < b[1]) return -1;
           else if(a[1] > b[1]) return 1;
           return 0;
        }
   }
   public static String[] F(BufferedReader bf) throws Exception
    {
        return (bf.readLine().split(" "));
    }
    
   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 double[] toDArray(String[] m) 
    { 
        double[] p=new double[m.length];
        for(int o=0;o<m.length;o++)
        {
            p[o]= Double.parseDouble(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 gcd(long a, long b)
        {
          if (b == 0)
            return a;
          return gcd(b, a % b); 
        }
     
        static long pow(long x, long y, long p)
      {
        if(y == 0) return 1l;
        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 int nCr(int n, int r, int p)
    {
        if (r > n - r)
            r = n - r;
 
        // The array C is going to store last
        // row of pascal triangle at the end.
        // And last entry of last row is nCr
        int C[] = new int[r + 1];
 
        C[0] = 1; // Top row of Pascal Triangle
 
        // One by constructs remaining rows of Pascal
        // Triangle from top to bottom
        for (int i = 1; i <= n; i++) {
 
            // Fill entries of current row using previous
            // row values
            for (int j = Math.min(i, r); j > 0; j--)
 
                // nCj = (n-1)Cj + (n-1)C(j-1);
                C[j] = (C[j] + C[j - 1]) % p;
        }
        return C[r];
    }
    // F(o,-1,tm,vis,par,o,mat);
   public static int log(int x)
   {
       boolean bol = true;

       if(((x-1)&x) == 0) bol=false;

       int c=0;
       while(x > 0)
       {
        x=(x/2);
        c++;
       }

       if(!bol) c--;
       return c;

   }

   
   static class sor implements Comparator<long[]>
   {
      public int compare(long[] a,long[] b)
      {
         if(a[0] == b[0])
         {
            if(a[1] < b[1]) return -1;
            return 1;
         }
         if(a[0] > b[0]) return 1;
         return -1;
      }
   }


    static long ncr(int n, int r)
    {
 
        // p holds the value of n*(n-1)*(n-2)...,
        // k holds the value of r*(r-1)...
        long p = 1, k = 1;
        if(n<r) return 0l;
 
        // C(n, r) == C(n, n-r),
        // choosing the smaller value
        if (n - r < r) {
            r = n - r;
        }
 
        if (r != 0) {
            while (r > 0) {
                p *= n;
                k *= r;
 
                // gcd of p, k
                long m = __gcd(p, k);
 
                // dividing by gcd, to simplify
                // product division by their gcd
                // saves from the overflow
                p /= m;
                k /= m;
 
                n--;
                r--;
            }
 
            // k should be simplified to 1
            // as C(n, r) is a natural number
            // (denominator should be 1 ) .
        }
        else {
            p = 1;
        }
 
        // if our approach is correct p = ans and k =1
        return p;
    }
 
    static long __gcd(long n1, long n2)
    {

        if(n1==0l) return n2;
        if(n2==0l) return n1;
        if(n1==1l || n2==1l) return 1l;
       // long gcd = 1;
        if(n1 == n2)  return n1;
        if(n1>n2) return __gcd(n1%n2,n2);
        return __gcd(n1,n2%n1);
        
    }
    static long big = -1000000000000000l;
    public static long F(int i,int k,int j,Long[][][] dp,long[] arr)
    {
      int m = dp[0].length-1;
        if( i >= arr.length)
        {
            if((k==1 && j==1) || k==0) return 0l;
            return big;
        }
        if(k < 0) return big;
        if(k == 0) return 0l;
        if(dp[i][k][j] != null)  return dp[i][k][j];
        long v1,v2;
        v1 = big;
        if(j == 0)
        {
            
             v1 = F(i+1,k,1,dp,arr);
             if(v1 > big) v1 += (arr[i]*(m+1l-k+0l));
             else v1=big;
             v1 = Math.max(v1,F(i+1,k,0,dp,arr));
             
        }
        else
        {
           
            v1 = F(i+1,k,1,dp,arr);
             if(v1 > big) v1 += (arr[i]*(m+1l-k+0l));
             else v1=big;
             v1 = Math.max(v1,F(i,k-1,0,dp,arr));
        }

        dp[i][k][j] = v1;
        return v1;


        

    }
    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 yy=1;//inInt(bf.readLine());
       


        
        for(int w=0;w<yy;w++)
        {
            // String str = bf.readLine();
          out.flush();
         
            String[] xlp = bf.readLine().split(" ");;;;;;
            //String st = bf.readLine();
        int n;//boolean bol=false;
           int m;//long a,b,c;
           int ioo;
           int  k;//pr(out,"vbc");
         n=inInt(xlp[0]);//m=inInt(xlp[1]);//int ta=inInt(xlp[2]);int tb=inInt(xlp[3]);k=inInt(xlp[4]);
        // k=inInt(xlp[2]);
          long[] a = toLArray(F(bf));
         
        m = inInt(bf.readLine());

        long[]  b = toLArray(F(bf));

        Arrays.sort(a);Arrays.sort(b);

        long[] sum = new long[n];
        long[] arr = new long[m];
        long s = 0l;

        for(int i=0;i<n;i++)
        {
            s += a[i];
            sum[i] = s;
        }

        long p =0l;
        for(int j=0;j<m;j++)
        {
            p+=b[j];
            arr[j]=p;
        }
        long res=-1000000000000l;
        long a1,a2;
        a1=a2=-1l;
        TreeMap<Long,Integer> map=new TreeMap<>();
        TreeMap<Long,Integer> map1=new TreeMap<>();

        for(int i=0;i<m;i++)
        {
            map.put(b[i],i);
        }
        for(int i=0;i<n;i++)
        {
            map1.put(a[i],i);
        }
        

        for(int i=0;i<n;i++)
        {
            int op=i;
            long  va= a[i]+1;
            Long in = map.floorKey(va);
            int index=-1;
            i = map1.get(map1.floorKey(va));
            if(in != null)
            {
                  index = map.get(in);
                  
            }
            long s1,s2;
            s1 = (i+1)*2l + 3l*(n-i-1l);
            s2 = (index+1l)*2l + 3l*(m-index-1l);
            long su = (s1-s2);
            if(su > res)
            {
                  res=su;
                  a1=s1;a2=s2;
            }
            else if(su == res)
            {
                  if(s1>a1)
                  {
                        a1=s1;a2=s2;
                  }
            }
            i=op;

        }
        {
            int i = -1;
            long  va= a[0]-1;
            Long in = map.floorKey(va);
            int index=-1;
            if(in != null)
            {
                  index = map.get(in);
                  
            }
            long s1,s2;
            s1 = (i+1)*2l + 3l*(n-i-1l);
            s2 = (index+1l)*2l + 3l*(m-index-1l);
            long su = (s1-s2);
            if(su > res)
            {
                  res=su;
                  a1=s1;a2=s2;
            }
            else if(su == res)
            {
                  if(s1>a1)
                  {
                        a1=s1;a2=s2;
                  }
            }
        }
        map.clear();
        map1.clear();
        {
        for(int i=0;i<n;i++)
        {
            map.put(a[i],i);
        }
        for(int i=0;i<m;i++)
        {
            map1.put(b[i],i);
        }

        for(int i=0;i<m;i++)
        {
            int op=i;
            long  va= b[i]+1;
            Long in = map.floorKey(va);
            i = map1.get(map1.floorKey(va));
            int index=-1;
            if(in != null)
            {
                  index = map.get(in);
                  
            }
            long s1,s2;
            s2 = (i+1)*2l + 3l*(m-i-1l);
            s1 = (index+1l)*2l + 3l*(n-index-1l);
            long su = (s1-s2);
            if(su > res)
            {
                  res=su;
                  a1=s1;a2=s2;
            }
            else if(su == res)
            {
                  if(s1>a1)
                  {
                        a1=s1;a2=s2;
                  }
            }
            i=op;

        }
        {
            int i = -1;
            long  va= b[0]-1;
            Long in = map.floorKey(va);
            int index=-1;
            if(in != null)
            {
                  index = map.get(in);
                  
            }
            long s1,s2;
            s2 = (i+1)*2l + 3l*(m-i-1l);
            s1 = (index+1l)*2l + 3l*(n-index-1l);///
            long su = (s1-s2);
            if(su > res)
            {
                  res=su;
                  a1=s1;a2=s2;
            }
            else if(su == res)
            {
                  if(s1>a1)
                  {
                        a1=s1;a2=s2;
                  }
            }
        }
      }
         
      pr(out,a1+":"+a2);
         

          
          




      

         
         
         



         
         
   }
    out.close();
   bf.close();
}}


/*
5 10 20 10 3
7 9 18 19 20
3 6 7 8 9 10 14 15 16 17


Kickstart
 String rp;
            rp = "Case #"+(w+1)+": "+(n-ans)+" ";
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); 
        }

Geometric median

public static double F(double[] x,double[] w)
    {
        double d1,d2;
        double S=0.00;
        for(double dp : w) S += dp;
        int k = 0;
        double sum = S - w[0]; // sum is the total weight of all `x[i] > x[k]`

        while(sum > S/2)
        {
            ++k;
            sum -= w[k];
        }
        d1=x[k];
        return d1;
        k = w.length-1;
       sum = S - w[k]; // sum is the total weight of all `x[i] > x[k]`

        while(sum > S/2)
        {
            --k;
            sum -= w[k];
        }
        d2=x[k];
        return new double[]{d1,d2};
    }
*/

















