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>
using namespace std;

typedef long long ll;

const long long MOD = 1000000007;
long long modpow(long long a, long long n, long long m)
{
    long long ans = 1;
    while (n)
    {
        if (n & 1)
        {
            ans = (ans * a) % m;
        }
        a = (a * a) % m;
        n >>= 1;
    }
    return ans;
}
long long combi(long long n, long long a)
{
    long long ans = 1, ans1 = 1;
    for (long long i = n - a + 1; i <= n; i++)
    {
        ans *= i % MOD;
        ans %= MOD;
    }

    for (long long i = 2; i <= a; i++)
        ans1 = (ans1 * i) % MOD;
    ans1 = modpow(ans1, MOD - 2, MOD);
    return ((ans % MOD) * ans1) % MOD;
}

void dfs(string s, char mx, ll N)
{
    if (s.length() == N)
    {
        cout << s.c_str() << endl;
    }
    else
    {
        for (char c = 'a'; c <= mx; c++)
        {
            dfs(s + c, ((c == mx) ? (char)(mx + 1) : mx), N);
        }
    }
}

int bfs(int a, int b, int h, int w, char tmp[][101])
{
    int ans[101][101] = {0};
    char c[101][101] = {0};
    queue<pair<int, int>> s;

    for (int i = 0; i <= h; i++)
    {
        for (int j = 0; j <= w; j++)
        {
            c[i][j] = tmp[i][j];
        }
    }

    s.push(make_pair(a, b));

    while (s.size() > 0)
    {

        pair<int, int> tmp = s.front();
        s.pop();
        c[tmp.first][tmp.second] = '#';
        for (int i = -1; i <= 1; i++)
        {
            for (int j = -1; j <= 1; j++)
            {
                if (tmp.first + i < 1 || tmp.first + i > h)
                {
                    continue;
                }
                if (tmp.second + j < 1 || tmp.second + j > w)
                {
                    continue;
                }
                if (i != 0 && j != 0)
                {
                    continue;
                }
                if (i == 0 && j == 0)
                {
                    continue;
                }
                if (c[tmp.first + i][tmp.second + j] == '#')
                {
                    continue;
                }
                c[tmp.first + i][tmp.second + j] = '#';
                if (ans[tmp.first + i][tmp.second + j] == 0)
                {
                    ans[tmp.first + i][tmp.second + j] = ans[tmp.first][tmp.second] + 1;
                }
                else
                {
                    ans[tmp.first + i][tmp.second + j] = min(ans[tmp.first + i][tmp.second + j], ans[tmp.first][tmp.second] + 1);
                }
                s.push(make_pair(tmp.first + i, tmp.second + j));
            }
        }
    }

    int asd = 0;

    for (int i = 1; i <= h; i++)
    {
        for (int j = 1; j <= w; j++)
        {
            asd = max(a, ans[i][j]);
        }
    }

    return asd;
}

int main()
{
    ll A, B, M;

    cin >> A >> B >> M;

    vector<ll> a(A);
    vector<ll> b(B);

    ll a_min = 100001;
    for(int i=0; i<A; i++){
        cin >> a[i];
        a_min = min(a_min, a[i]);
    }
    ll b_min = 100001;
    for(int i=0; i<B; i++){
        cin >> b[i];
        b_min = min(b_min, b[i]);
    }

    ll ans = a_min + b_min;
    for(int i=0; i<M; i++){
        ll x, y, m;
        cin >> x >> y >> m;
        ans = min(ans, a[x-1]+b[y-1]-m);
    }

    cout << ans << endl;
    return 0;
}