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: Accepted)

#include <bits/stdc++.h>
#define _GLIBCXX_DEBUG
#define INF 100000000
using namespace std;
const int mod = 1000000007;

int main() {

  int A,B,M,mina=INF,minb=INF,ans=INF;
  cin>>A>>B>>M;
  vector<int> a(A),b(B);
  for(int i=0; i<A; i++){cin>>a[i];mina=min(mina,a[i]);}
  for(int i=0; i<B; i++){cin>>b[i];minb=min(minb,b[i]);}
  vector<vector<int>> m(M,vector<int>(3));
  ans=mina+minb;
  for(int i=0; i<M; i++){
    for(int j=0; j<3; j++){
      cin>>m[i][j];
    }
      ans=min(ans,a[m[i][0]-1]+b[m[i][1]-1]-m[i][2]);
  }
  //sort(A.begin(),A.end());
  //for (int i = 0; i < N; i++) { maximum = max(maximum , A[i]);}
  //for (int i = 0; i < N; i++) { minimum = min(minimum , A[i]);}
  cout << ans << endl;
}