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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

int main(){
    int A, B ,M, i,j;
    cin >> A;
    cin >> B;
    cin >> M;
    int a[100000], b[100000];
    long long c_m[100000][100000],l;
  
    for(int i=0;i<A;i++){
        cin >> j;
        a[i] = j;
    }
    for(int i=0;i<B;i++){
        cin >> j;
        b[i] = j;
    }
  	for(int i = 0;i<A;i++){
      for(int j = 0;j<B;j++){
        c_m[i][j] = 0; 
      }
    }

    for(int k=0; k<M;k++){
      	cin >> i;
      	cin >> j;
      	cin >> l;
		if(c_m[i-1][j-1]<l){
          c_m[i-1][j-1] = l;
        }
    }

    int min_money = 1000000;
    int money;
  
  	for(int i=0;i<A;i++){
    	for(int j=0; j<B;j++){
            money = a[i] + b[j] - c_m[i][j]; 
            chmin(min_money,money);      	
        }
    }

    cout << min_money << endl;

}