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: Wrong Answer)

#include<iostream>
#include<iomanip>
#include<limits>
#include<string>
#include<vector>
#include<algorithm>
#include<stdio.h>
#include<cmath>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;

#define big ((ll)(1e9)+7)

int main(){
  int a,b,m;
  cin >> a >> b >> m;
  vector<int> A;
  vector<int> B;
  for(int i = 0;i < a;i++)
  {
    int p;
    cin >> p;
    A.push_back(p);
  }
  for(int i = 0;i < b;i++)
  {
    int p;
    cin >> p;
    B.push_back(p);
  }
  vector<int> x;
  vector<int> y;
  vector<int> C;
  for(int i = 0;i < m;i++)
  {
    int p,q,r;
    cin >> p >> q >> r;
    x.push_back(p);
    y.push_back(q);
    C.push_back(r);
  }

  vector<ll> ans;
  for(int i = 0;i < a;i++)
  {
    for(int j = 0;j < b;j++)
    {
      int Ans = 0;
      if(m == 1)
      {
        if(x[0] == i+1 && y[0] == j+1)
        {
          Ans -= C[0];
        }
      }
      else
      {
        for(int l = 0;l < m-1;l++)
        {
          cout << x[l] << " " << y[l] << endl;
          if(x[l] == i+1 && y[l] == j+1)
          {
            if(x[l] == x[l+1] && y[l] == y[l+1])
            {
              Ans -= C[l+1];
              // cout << Ans << endl;
              break;
            }
            else
            {
              Ans -= C[l];
              break;
            }
          }
        }
      }
      Ans += A[i]+B[j];
      ans.push_back(Ans);
    }
  }
  // for(int i = 0;i < ans.size();i++)
  // {
  //   cout << ans[i] << " ";
  // }
  sort(ans.begin(),ans.end());
  // cout << endl;
  cout << ans[0] << endl;
}