Problem:
Score : 200 points

Problem Statement

You are visiting a large electronics store to buy a refrigerator and a microwave.

The store sells A kinds of refrigerators and B kinds of microwaves. The i-th refrigerator ( 1 \le i \le A ) is sold at a_i yen (the currency of Japan), and the j-th microwave ( 1 \le j \le B ) is sold at b_j yen.

You have M discount tickets. With the i-th ticket ( 1 \le i \le M ), you can get a discount of c_i yen from the total price when buying the x_i-th refrigerator and the y_i-th microwave together. Only one ticket can be used at a time.

You are planning to buy one refrigerator and one microwave. Find the minimum amount of money required.

Constraints

All values in input are integers.

1 \le A \le 10^5

1 \le B \le 10^5

1 \le M \le 10^5

1 \le a_i , b_i , c_i \le 10^5

1 \le x_i \le A

1 \le y_i \le B

c_i \le a_{x_i} + b_{y_i}

Input

Input is given from Standard Input in the following format:

A B M
a_1 a_2 ... a_A
b_1 b_2 ... b_B
x_1 y_1 c_1
\vdots
x_M y_M c_M

Output

Print the answer.

Sample Input 1

2 3 1
3 3
3 3 3
1 2 1

Sample Output 1

5

With the ticket, you can get the 1-st refrigerator and the 2-nd microwave for 3+3-1=5 yen.

Sample Input 2

1 1 2
10
10
1 1 5
1 1 10

Sample Output 2

10

Note that you cannot use more than one ticket at a time.

Sample Input 3

2 2 1
3 5
3 5
2 2 2

Sample Output 3

6

You can get the 1-st refrigerator and the 1-st microwave for 6 yen, which is the minimum amount to pay in this case.
Note that using a ticket is optional.

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

require 'set'

def main
    a,b,m = gets.chomp.split(" ").map(&:to_i)
    as = gets.chomp.split(" ").map(&:to_i)
    bs = gets.chomp.split(" ").map(&:to_i)
    ds = m.times.map{ gets.split(" ").map(&:to_i) }

    as.sort!
    bs.sort!
    ans = as[0] + bs[0]
    (0..m-1).each do |i|
        price = as[ds[i][0]-1] + bs[ds[i][1]-1] - ds[i][2]
        ans = price if ans > price
    end
    puts ans
end

main

## イディオムなど
# until s.empty? do
#
## 配列のインデックス
# s[0,6] # 0番目から6文字
# s[0..6] # 0番目から6番目まで
# s[5..-1] # 5番目から最後まで
#
## 全探索
# [0, 1].repeated_permutation(n) do |combination| # bit全探索
# array.combination(i) do |selected| # array から要素が i 個である部分列を全部抽出
#
## 無向グラフの幅優先探索(BFS)
# to = Array.new(n).map{Array.new()} # ある頂点と繋がる頂点のリスト
# m.times do
#     a,b = gets.chomp.split(" ").map(&:to_i)
#     to[a].push b
#     to[b].push a
# end
# q = Queue.new # BFS は queue
# ds = Array.new(n, -1)
# pre = Array.new(n, -1)
#
## その他
# a1 = Marshal.load(Marshal.dump(a2)) # 深いコピー
# matrix = Array.new(n).map{Array.new()} # 二次元配列の初期化
#