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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
int main() {
    int a, b, m, min =10000000;
    int* an;
    int* bn;
    scanf("%d%d%d", &a, &b, &m);
    an = (int*)malloc(sizeof(int) * (a + 1));
    bn = (int*)malloc(sizeof(int) * (b + 1));

    for (int i = 1; i <= a; i++) { scanf("%d", &an[i]); }
    for (int i = 1; i <= b; i++) { scanf("%d", &bn[i]); }


    for (int i = 0; i < m; i++) {
        int d1, d2, d3;
        scanf("%d%d%d", &d1, &d2, &d3);
        int cost = an[d1] + bn[d2] - d3;
        if (cost < min) { min = cost; }
    }
    int bmin = bn[1],amin=an[1];
    for (int i = 2; i <= b; i++) {
        if (bn[i] < bmin) { bmin = bn[i]; }
    }
    for (int i = 2; i <= a; i++) {
        if (an[i] < amin) { amin = an[i]; }
    }

    if (min > amin + bmin) { min = amin + bmin; }

    printf("%d", min);
    return 0;
}