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 Lua (Lua 5.3.5) (judge verdict: Accepted)

--url:https://gist.github.com/LukeMS/89dc587abd786f92d60886f4977b1953
--[[  Priority Queue implemented in lua, based on a binary heap.
Copyright (C) 2017 Lucas de Morais Siqueira <lucas.morais.siqueira@gmail.com>
License: zlib
  This software is provided 'as-is', without any express or implied
  warranty. In no event will the authors be held liable for any damages
  arising from the use of this software.
  Permission is granted to anyone to use this software for any purpose,
  including commercial applications, and to alter it and redistribute it
  freely, subject to the following restrictions:
  1. The origin of this software must not be misrepresented; you must not
     claim that you wrote the original software. If you use this software
     in a product, an acknowledgement in the product documentation would be
     appreciated but is not required.
  2. Altered source versions must be plainly marked as such, and must not be
     misrepresented as being the original software.
  3. This notice may not be removed or altered from any source distribution.
]]--

local floor = math.floor


local PriorityQueue = {}
PriorityQueue.__index = PriorityQueue

setmetatable(
    PriorityQueue,
    {
        __call = function (self)
            setmetatable({}, self)
            self:initialize()
            return self
        end
    }
)


function PriorityQueue:initialize()
    --[[  Initialization.
    Example:
        PriorityQueue = require("priority_queue")
        pq = PriorityQueue()
    ]]--
    self.heap = {}
    self.current_size = 0
end

function PriorityQueue:empty()
    return self.current_size == 0
end

function PriorityQueue:size()
    return self.current_size
end

function PriorityQueue:swim()
    -- Swim up on the tree and fix the order heap property.
    local heap = self.heap
    local floor = floor
    local i = self.current_size

    while floor(i / 2) > 0 do
        local half = floor(i / 2)
        if heap[i][2] < heap[half][2] then
            heap[i], heap[half] = heap[half], heap[i]
        end
        i = half
    end
end

function PriorityQueue:put(v, p)
    --[[ Put an item on the queue.
    Args:
        v: the item to be stored
        p(number): the priority of the item
    ]]--
    --

    self.heap[self.current_size + 1] = {v, p}
    self.current_size = self.current_size + 1
    self:swim()
end

function PriorityQueue:sink()
    -- Sink down on the tree and fix the order heap property.
    local size = self.current_size
    local heap = self.heap
    local i = 1

    while (i * 2) <= size do
        local mc = self:min_child(i)
        if heap[i][2] > heap[mc][2] then
            heap[i], heap[mc] = heap[mc], heap[i]
        end
        i = mc
    end
end

function PriorityQueue:min_child(i)
    if (i * 2) + 1 > self.current_size then
        return i * 2
    else
        if self.heap[i * 2][2] < self.heap[i * 2 + 1][2] then
            return i * 2
        else
            return i * 2 + 1
        end
    end
end

function PriorityQueue:pop()
    -- Remove and return the top priority item
    local heap = self.heap
    local retval = heap[1][1]
    heap[1] = heap[self.current_size]
    heap[self.current_size] = nil
    self.current_size = self.current_size - 1
    self:sink()
    return retval
end

--return PriorityQueue

local x,y,a,b,c=io.read("n","n","n","n","n")
local q=PriorityQueue()
local t={}

for i=1,a do
    local input=io.read("n")
    q:put(input,-input)
end
for i=1,x do
    table.insert(t,q:pop())
end

q:initialize()

for i=1,b do
    local input=io.read("n")
    q:put(input,-input)
end
for i=1,y do
    table.insert(t,q:pop())
end

for i=1,c do
    table.insert(t,io.read("n"))
end

table.sort(t,function(a,b)
    return a>b
end)

local max=0
for i=1,x+y do
    max=max+t[i]
end
print(max)