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
#include <bits/stdc++.h>
/*
#include<boost/multiprecision/cpp_int.hpp>
namespace mp=boost::multiprecision;
*/
using namespace std;
#define ll  int
#define lo mp::cpp_int
#define pb push_back
#define pob pop_back
#define mk make_pair
#define endl "\n"
#define ce(x) cout << x <<"\n"
#define fill(s,v) memset(s,v,sizeof(s))
#define speed  { ios_base::sync_with_stdio(false); cin.tie(0); }
#define deb(x) cout<<#x<<" "<<x<<endl; 
#define INFINITE 1000000009
#define ull unsigned long long int
#define take(a,n) for(ll pp=0;pp<n;pp++) { cin>>a[pp]; }
#define take2(a,n) for(ll p=0;p<n;p++) { ll temp;cin>>temp;a.pb(temp); }
#define f(i,n) for(ll i=0;i<n;i++)
#define f1(i,a,n) for(ll i=a;i<n;i++)
#define show(a,n) for(ll p=0;p<n;p++) cout<<a[p]<<" ";
#define _ <<" "<<
#define nl cout<<endl;
#define sq(x) ((x)*(x))
#define call(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define ff first
#define ss second
#define lld long double
#define sc(n) scanf("%lld",&n);
#define de2(x,y) cout<<#x<<" = "<<(x) <<' '<< #y<<" = "<<y<<endl;typedef pair<int, int> pii;
#define copy(b,a,n) for(ll i=0;i<n;i++) b[i]=a[i];
#define clock() clock_out()
#define tc(tt) \
    ll tt;    \
    cin >> tt; \
    for (int _tt = 0; _tt < tt; _tt++) 
typedef vector<vector<ll>> matrix ;  
typedef pair<ll,ll> pii;
typedef vector<pii> vii;
typedef vector<ll> vi;   
const ll dx[4] = {-1, 0, 1, 0};
const ll dy[4] = {0, -1, 0, 1};
const ll mod = 1000000007;
ll gcd(ll a,ll b){if(!b)return a;return gcd(b,a%b);} 
ll power_mod(ll x,ll y,ll p){ll res=1;x%=p;while(y>0){if(y&1)res=(res*x)%p;y=y>>1;x=(x*x)%p;}return res;} 
ll modexpo(ll x,ll n,ll M){if(n==0)return 1;else if(n%2 == 0)return modexpo((x*x)%M,n/2,M);else return(x*modexpo((x*x)%M,(n-1)/2,M))%M;}
ll lcm(ll a, ll b){return (a*b)/gcd(a, b);}
ll sumofdigits(ll n){ll out=0;while(n){out+=(n%10);n/=10;}return out;}
bool isPrime(ll n){if (n <= 1)  
    return false; if (n <= 3)  return true; if (n%2 == 0 || n%3 == 0) return false; 
    for (ll i=5; i*i<=n; i=i+6) if (n%i == 0 || n%(i+2) == 0) return false;return true;} 
// ll power(ll a,ll b){if(b==0)return 1;if(b&1)return a*power(a,b-1);ll res=power(a,b/2);return res*res;}
ll bexpo(ll x,ll n){if(n==0)return 1;else if(n%2==0)return bexpo(x*x,n/2);else return x*bexpo(x*x,(n-1)/2);}
ll log(ll d){return log2(d);}
bool isPower(ll x, ll y){if(x==1) return (y==1); ll pow=1; while(pow<y) pow*=x; return(pow == y);}  
bool isPerfectSquare(long double x){long double sr = sqrt(x);return ((sr - floor(sr)) == 0);}   
bool compare(const pair<ll,ll> &a, const pair<ll,ll> &b) { return (a.second < b.second);} 
ll isPalindrome(string s){ll l=0;ll h=s.size()-1;while(h>l){if(s[l++]!=s[h--])return 0;}return 1;} 
ll setbits(ll n) { if(n==0)return 0; else return 1+setbits(n&(n-1)); } 
ll modInverse(ll A,ll M){return modexpo(A,M-2,M);}
ll suma(ll *a,ll n){ll sum=0;for(ll i=0;i<n;i++)sum+=a[i];return sum;}
// void clock_out(){
//   cerr<<"\nTime Elapsed : " << 1.0*clock() / CLOCKS_PER_SEC << " s\n";
// }
//------modular inverse by extended euclid-------//

// ll d, x, y;
// void extendedEuclid(ll A, ll B) {
//  if(B == 0) {
//  d = A;
//  x = 1;
//  y = 0;
//  }
//  else {
//  extendedEuclid(B, A%B);
//  ll temp = x;
//  x = y;
//  y = temp - (A/B)*y;
//  }
// }

// ll modInverse2(ll A, ll M){
// extendedEuclid(A,M);
// return (x%M+M)%M;
// }

//---------------------------------//

//---------------matrix exponentiation-------------------//

// matrix mul(matrix a,matrix b){
//   matrix res(n+1,vector<ll>(n+1));
//   f(i,n)f(j,n)f(k,n)res[i][j]=(res[i][j]%mod+((a[i][k]%mod)*(b[k][j]%mod))%mod)%mod;
//    return res;
// }
// matrix pow(matrix A,ll p) {
//     if(p==1) return A;
//     matrix res=pow(A,p/2) ;
//     res=mul(res,res);
//     if(p&1) res=mul(res,A) ;
//     return res;
// }

//-------------------------------------------------//

vector<ll> factorize(ll n) {
    vector<ll> res;
    for (ll i = 2; i * i <= n; ++i) {
        while (n % i == 0) {
            res.push_back(i);
            n /= i;
        }
    }
    if (n != 1) {
        res.push_back(n);
    }
    return res;
}

vector<ll> divisors(ll n){ 
    vector<ll> res;
    for (ll i=1; i<=sqrt(n); i++){ 
        if (n%i == 0){ if (n/i == i){ res.pb(i); } else{ res.pb(n/i); res.pb(i); } } 
    } 
    return res;
}   
vector<ll> primeFactors(ll n){  
    if(n==1) return vector<ll>(1,1);
    bool idx[1]={false};
    vector<ll> res;
    while (n%2 == 0){  
        if(!idx[1]){idx[1]=true;res.pb(2);}
        n = n/2;  
    }  
    for (ll i=3;i*i<=n; i+=2){
        bool once=false;
        while (n%i==0){ if(!once){res.pb(i);once=true;} n=n/i; }  
    }  
    if (n > 2)res.pb(n);
    return res;
}

//--------------------------------------------//

ll dxx[] = {1, 1, 0, -1, -1, -1, 0, 1};
ll dyy[] = {0, 1, 1, 1, 0, -1, -1, -1};

//----check if a point is in range-----///
///__builtin_ctz(x) --> get power of 2 dude !!!! 

bool check(ll x,ll y,ll n,ll m){return(x>=0&&x<n&&y>=0&&y<m);}

const int FULL_MASK = (1 << 10) - 1;
#define pi 3.1415926535897932

vector<vector<ll>> ed;

void bfs(ll s, vector<ll> &d) {
    d[s] = 0;
    queue<ll> q;
    q.push(s);
    while (!q.empty()) {
        ll v = q.front();
        q.pop();
        for (auto to : ed[v]) {
            if (d[to] == 1e9) {
                d[to] = d[v] + 1;
                q.push(to);
            }
        }
    }
} 
// void dfs(int u){
//     vis[u]=1;
//     // temp.push_back(u);
//     for(int i=0;i<adj[u].size();i++){
//         if(!vis[adj[u][i]])
//             dfs(adj[u][i]);
//     }
// }
ll maxSubArraySum(ll a[],ll size) 
{ 
    ll max_so_far = a[0]; 
    ll curr_max = a[0]; 
  
   for (ll i = 1; i < size; i++) 
   { 
        curr_max = max(a[i], curr_max+a[i]); 
        max_so_far = max(max_so_far, curr_max); 
   } 
   return max_so_far; 
} 

struct node{
    ll a,b,c,d;
};

bool c(ll mid,ll n,ll m,ll k){
    ll less=0;
    for(ll i=1;i<=n;i++){
        less+=min(m,(mid)/i);
    }
    if(less>=k){
        return true;
    }
    else{
        return false;
    }
}

int a[223456], b[223456], A, B, res, cur, ans;
map<int, int> t;
 
void solve(){
        int n,m;
        cin>>n;
        for(int i=0;i<n;++i){
                cin>>a[i];
                ++t[a[i]];
        }
        cin>>m;
        for(int i=0;i<m;++i){
                cin>>b[i];
                --t[b[i]];
        }
        cur=res=3*(n-m);
        for(auto p:t){
                cur-=p.second;
                if(res<cur){
                        res=cur;
                        ans=p.first;
                }
        }
        for(int i=0;i<n;++i)
                if(a[i]<=ans)A+=2;
                else A+=3;
        for(int i=0;i<m;++i)
                if(b[i]<=ans)B+=2;
                else B+=3;    
        cout<<A<<":"<<B;
}
int main(){
    speed
    // tc(tt){
        solve();
    // }
}
/*


*/