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 Python (3.4.3) (judge verdict: Runtime Error)

import sys
sys.setrecursionlimit(10**6)

# 組み合わせ
mod = 10 ** 9 + 7
N = 2* 10 ** 5  # N は必要分だけ用意する
fact = [1, 1]  # fact[n] = (n! mod p)
factinv = [1, 1]  # factinv[n] = ((n!)^(-1) mod p)
inv = [0, 1]  # factinv 計算用

for i in range(2, N + 1):
    fact.append((fact[-1] * i) % mod)
    inv.append((-inv[mod % i] * (mod // i)) % mod)
    factinv.append((factinv[-1] * inv[-1]) % mod)

def cmb(n, r):
    if (r < 0) or (n < r):
        return 0
    r = min(r, n - r)
    return fact[n] * factinv[r] * factinv[n-r] % mod
  
# 深さ優先探索
def dfs(v, p=-1):

    # print(v, p, [x + 1 for x in E[v]])
    
    dp_l = 1  # 自分自身しかない場合
    size_l = 1 # 自分自身のサイズを含める

    # 子の情報をあわせて数え上げる
    for u in E[v]:
        
        # 親へのルートは辿らない
        if u == p:
            continue

        d, s = dfs(u, v)
        d %= mod
        size_l += s
        dp_l = dp_l * d * cmb(size_l-1, s) % mod

    dp[v] = dp_l
    size[v] = size_l
        
    return (int(dp_l), size_l)
  
def bfs(v, p=-1):
    
    for u in E[v]:
        if u == p:
            continue
        
        # u がなくなった場合の dp[v] 値、 size を求める
        # 一旦、uを切り離す

        partial_v = int(dp[v] / dp[u] / cmb(N - 1, size[u]))
        
        # u に v 側の部分木が接続された場合の値を求める
        # u に v を根とする部分木をつける

        dp_l = dp[u] * partial_v * cmb(N-1, size[u]-1)
        
        dp[u] = dp_l
    
        bfs(u,v)

# 入力処理
N = int(input())
E = [[] for _ in range(N)]

for _ in range(N-1):
    a, b = map(int, input().split())
    E[a-1].append(b-1)
    E[b-1].append(a-1)
      
dp = [1] * N
size = [1] * N


dfs(0)

bfs(0)

for i in range(N):
    print(dp[i])