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

#include<iostream>
#include<sstream>

using namespace std;


int main() {

	unsigned int A;
	unsigned int B;
	unsigned int M;
	unsigned int i;

	cin >> A >> B >> M;

	unsigned int* Ax = new unsigned int[A];
	unsigned int* Bx = new unsigned int[B];

	for (i = 0; i < A; i++) {

		cin >> Ax[i];

	}

	for (i = 0; i < B; i++) {

		cin >> Bx[i];

	}

	unsigned int* Ma = new unsigned int[M];
	unsigned int* Mb = new unsigned int[M];
	unsigned int* Mc = new unsigned int[M];

	for (i = 0; i < M; i++) {

		cin >> Ma[i] >> Mb[i] >> Mc[i];

	}


	unsigned int j;
	long int ans1;
	long int ans2;

	ans2 = 1000000000;


	for (i = 0; i < M; i++) {

		ans1 = Ax[Ma[i] - 1] + Bx[Mb[i] - 1] - Mc[i];

		ans2 > ans1 ? ans2 = ans1 : ans2 = ans2;
	}
	int AA = 10000000;
	for (i = 0; i < A; i++) {
		(AA > Ax[i]) ? AA = Ax[i] : AA;
	}

	int BB = 10000000;
	for (i = 0; i < B; i++) {
		(BB > Bx[i]) ? BB = Bx[i] : BB;
	}
	ans1 = AA + BB;
	ans2 > ans1 ? ans2 = ans1 : ans2 = ans2;

	delete[] Ma;
	delete[]Mb;
	delete[]Mc;
	delete[]Ax;
	delete[]Bx;


	cout << ans2 << endl;

	return 0;


}