Problem:
Score : 500 points

Problem Statement

You are going to eat X red apples and Y green apples.

You have A red apples of deliciousness p_1,p_2, \dots, p_A, B green apples of deliciousness q_1,q_2, \dots, q_B, and C colorless apples of deliciousness r_1,r_2, \dots, r_C.

Before eating a colorless apple, you can paint it red or green, and it will count as a red or green apple, respectively.

From the apples above, you will choose the apples to eat while making the sum of the deliciousness of the eaten apples as large as possible.

Find the maximum possible sum of the deliciousness of the eaten apples that can be achieved when optimally coloring zero or more colorless apples.

Constraints

1 \leq X \leq A \leq 10^5

1 \leq Y \leq B \leq 10^5

1 \leq C \leq 10^5

1 \leq p_i \leq 10^9

1 \leq q_i \leq 10^9

1 \leq r_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X Y A B C
p_1 p_2 ... p_A
q_1 q_2 ... q_B
r_1 r_2 ... r_C

Output

Print the maximum possible sum of the deliciousness of the eaten apples.

Sample Input 1

1 2 2 2 1
2 4
5 1
3

Sample Output 1

12

The maximum possible sum of the deliciousness of the eaten apples can be achieved as follows:

Eat the 2-nd red apple.

Eat the 1-st green apple.

Paint the 1-st colorless apple green and eat it.

Sample Input 2

2 2 2 2 2
8 6
9 1
2 1

Sample Output 2

25

Sample Input 3

2 2 4 4 4
11 12 13 14
21 22 23 24
1 2 3 4

Sample Output 3

74

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>//数据输入输出流
#include <cstring>//字符串操作函数
#include <cstdio>//C的输入输出
#include <cstdlib>//定义杂项函数及内存分配函数
#include <cmath>//C中的数学函数
#include <string>//c++中的string类 他不能用strcpy等c函数去操作
#include <vector>//STL vetor
#include <list>//STL list
#include <map>// STL map
#include <queue>// STL queue
#include <stack>//sTL stack
#include <bitset>//bitset可按位定义串
//比如：bitset <1000> all;定义一个1000位的串
#include <algorithm>//STL各种算法 比如 swap sort merge max min 比较
#include <numeric>//常用数字操作 一般和algorithm搭配使用
#include <functional>//STL定义运算函数（代替运算符）
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define LOCAL
#define zeros(a,n) memset(a,0,(n)*sizeof(a[0]))
const int modn = 1e9+7;
int mod(int x) { return x<0?x+modn:x<modn?x:x-modn; }
const int N = 200005;
int p[N],q[N],e[N];
int n,x,y,a,b,c,idx;
int d[N], r[N];
bool cmp(const int& a, const int& b){
    return a>b;
}
int main()
{
    ios::sync_with_stdio(false); cin.tie(0);
    cin>>x>>y>>a>>b>>c;
    for(int i=0;i<a;++i)cin>>p[i];
    for(int i=0;i<b;++i)cin>>q[i];
    for(int i=0;i<c;++i)cin>>e[i];
    sort(p,p+a,cmp);
    sort(q,q+b,cmp);
    sort(e,e+c);
    idx = 0;
    priority_queue<int>que1,que2,que3;
    for(int i=0;i<min(x,a); ++i) que1.push(-p[i]);
    for(int j=0;j<min(y,b);++j) que2.push(-q[j]);
    for(int i=0;i<c;++i) que3.push(e[i]);
    while(que1.size()<x && que3.size()){
        que1.push(-que3.top());
        que3.pop();
    }
    while(que2.size()<y && que3.size()){
        que2.push(-que3.top());
        que3.pop();
    }
    while(que3.size()){
        int x = -que1.top();
        int y= -que2.top();
        int z = que3.top(); que3.pop();
        if(x > y){
            if(z > y) {
                que2.pop();
                que2.push(-z);
            }
        }else{
            if(z > x) {
                que1.pop();
                que1.push(-z);
            }
        }
    }
    LL ans = 0;
    while(que1.size()){
        ans += -que1.top();
        que1.pop();
    }
    while(que2.size()){
        ans += -que2.top();
        que2.pop();
    }
    /*
    for(int i=0;i<min(x,a);++i) d[idx++]=p[i];
    for(int i=0;i<min(b,y);++i) d[idx++]=q[i];
    sort(d,d+idx);
    priority_queue<int>que;
    for(int i=0, j=0; i<=min(c-1,idx-1), j<c;++i,++j){
        que.push(d[i]);
        que.push(e[j]);
    }
    LL ans=0;
    for(int i=c;i<=idx-1;++i) ans+=d[i];
    int t = (x + y) - (idx - c);
    while(t--){
        ans += que.top();
        que.pop();
    }
    */
    cout<<ans<<"\n"; 
    return 0;
}