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 Go (1.6) (judge verdict: Runtime Error)

// 赤X+緑Y個を選びたい、赤Aと緑BとCがありCは赤か緑に変更可能
// A+BでX+Yが足りる場合（条件から絶対足りるわコレ アホな誤読してた） Cで置換できるものがあれば置換する
// 置換はRの大きいものからP,Qに対してにぶたんして、優先順位の高いやつからやる
// 優先順位の高いやつ、それは何
// なんかどうせ使うんだしP,Qどっちにしても良さそうだしそうする
// これは懸念ですが雑にやったのでpopleftがおかしいし多分TLEなると思うこれ

package main

import (
	"fmt"
	"sort"
)

func main() {
	var X, Y, A, B, C int
	fmt.Scan(&X, &Y, &A, &B, &C)
	P, Q, R := make([]int, A), make([]int, B), make([]int, C)

	for i := 0; i < A; i++ {
		var V int
		fmt.Scan(&V)
		P[i] = V
	}
	for i := 0; i < B; i++ {
		var V int
		fmt.Scan(&V)
		Q[i] = V
	}
	for i := 0; i < C; i++ {
		var V int
		fmt.Scan(&V)
		R[i] = V
	}
	sort.Sort(sort.Reverse(sort.IntSlice(P)))
	sort.Sort(sort.Reverse(sort.IntSlice(Q)))
	sort.Sort(sort.Reverse(sort.IntSlice(R)))
	P = P[:X]
	Q = Q[:Y]
	// まずRをPに使えるだけ試し、次に余ったRをQに使えるだけ試す
	// いつもならlower boundをフルスクラッチで実装するんだけどこれたまたまsort使ってるのでsortパッケージから使ってみることにする

	usedCount := 0
	for _, r := range R {
		// fmt.Printf("target = %d\n", r)
		idx := sort.Search(X, func(i int) bool { return P[i] < r })
		// fmt.Printf("idx:= %d\n", idx)
		if idx >= X {
			// fmt.Println("もう置換不可能")
			break
		} else {
			// fmt.Printf("置換可能： P[%d]=%d\n", idx, P[idx])
			P[idx] = r
			usedCount += 1
		}
	}
	R = R[usedCount:]
	for _, r := range R {
		// fmt.Printf("target = %d\n", r)
		idx := sort.Search(X, func(i int) bool { return Q[i] < r })
		// fmt.Printf("idx:= %d\n", idx)
		if idx >= Y {
			// fmt.Println("もう置換不可能")
			break
		} else {
			// fmt.Printf("置換可能： Q[%d]=%d\n", idx, Q[idx])
			Q[idx] = r
			usedCount += 1
		}
	}
	ans := 0
	for _, p := range P {
		ans += p
	}
	for _, q := range Q {
		ans += q
	}
	fmt.Println(ans)
}