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: Time Limit Exceeded)

#include <iostream>
#include <stdio.h>
#include <vector>
#include <utility>
#include <string>
#include <algorithm>
#include <cmath>
#include <queue>
 
typedef long long ll;

#define rep(i,n) for(ll i = 0; i < n; i++)

using namespace std;
 
const ll INF = ll(1e18)+1;
const double PI=3.14159265358979323846;

int main(){
  int A,B,M,a[10100]={},b[100100]={},x[10100]={},y[10100]={},c[10100]={};
  cin >> A >> B >> M;
  rep(i,A) cin >> a[i+1];
  rep(i,B) cin >> b[i+1];  
  rep(i,M) cin >> x[i] >> y[i] >> c[i];

  int ans = 1e+9;
  rep(i,M){
    ans = min(ans,a[x[i]]+b[y[i]]-c[i]);
  }
  rep(i,A){
    rep(j,B){
      ans = min(ans,a[i+1]+b[i+1]);
    }    
  }
  cout << ans << endl;
}