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 asc(const void* n1,const void* n2)
{
  return *(int*)n1 - *(int*)n2;
}

int main(void)
{
  int A,B,M;
  int min = 1000000;
  scanf("%d%d%d",&A,&B,&M);

  int a[A+1],b[B+1],x[M+1],y[M+1],c[M+1];

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

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

  for(int i=1; i<=M; i++){
    scanf("%d",&x[i]);
    scanf("%d",&y[i]);
    scanf("%d",&c[i]);
  }

  for(int i=1; i<=M; i++){
    if(min > a[x[i]]+b[y[i]]-c[i]){
      min = a[x[i]]+b[y[i]]-c[i];
    }
  }

  qsort(a+1,A,sizeof(int),asc);
  qsort(b+1,A,sizeof(int),asc);  

  if(min > a[1]+b[1]){
    min = a[1]+b[1];
  }
//  printf("%d %d",a[1],b[1]);  



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

}