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)

#define ll long long int
#include <bits/stdc++.h>
//#include <ext/pb_ds/tree_policy.hpp>
//#include <ext/pb_ds/assoc_container.hpp>
using namespace std;

//using namespace __gnu_pbds;

#define         ff               first
#define         ss               second
#define         mp               make_pair
#define         pb               push_back
#define         pf               push_front
#define         vi               vector<ll>
#define         pii              pair<ll,ll>
#define         qi               queue<ll>
#define         mem(a,b)         memset(a,b,sizeof(a))
#define         print(a)         cout<<a<<'\n'
#define         sorted(arr)      sort(arr.begin(),arr.end())
#define         gcd(a,b)         gcd((a),(b))
#define         lcm(a,b)         ((a)*(b))/gcd((a),(b))
#define         all(v)           v.begin(),v.end()
#define         read(a)          scanf("%d", &a)
#define         f(a,n)           for(ll i=a;i<n;i++)
#define         in2(n,m,arr)     for(ll i=0;i<n;i++){for(ll j=0;j<m;j++)cin>>arr[i][j];}
#define         debug(a)         cerr<<a<<" ";
#define         debug1(a)        cerr<<a<<'\n'
 
/*typedef tree<int, null_type, less<int>, rb_tree_tag , tree_order_statistics_node_update> Set;*/
 
 
const unsigned ll mod = 1e9 + 7 ;
const ll inf = (ll)1e18 + 2;
const ll N = (ll)2e5 + 2;

ll power( ll x , ll y)
{
    ll res = 1;
    while (y > 0)
    {
        if (y & 1)
            res = (res * x) ;
        y = y >> 1;
        x = (x * x) ;
    }
    return res  ;
}

//struct node{
//    ll x , y , z;
//};
//map<pair<ll, ll>,ll> c;
//vector<set<ll>> tree;
//vector<ll> ar ;
//set<ll> s;
//ll get(ll x)
//{
//    if(x == ar[x])
//        return x;
//    return ar[x] = get(ar[x]);
//}
//void merge(ll a ,ll b)
//{
//    a = get(a);
//    b = get(b);
//    if(a == b)
//    {
//        return ;
//    }
//    ar[b] = ar[a];
// }

//ll sev[100006];
//vector<ll> pri;
//void seive()
//{
//    for(ll i = 2 ; i < 100006 ; i ++)
//    {
//        if(!sev[i])
//        {
//            for(ll j = i ; j < 100006 ; j += i)
//            {
//                if(!sev[j])
//                    sev[j] = i;
//            }
//        }
//        if(sev[i] == i)
//            pri.pb(i);
//    }
//}

void solve()
{
    ll a , b , m;
    cin >> a >> b >> m;
    ll arr[a] , brr[b];
    ll ans = 0,mn = inf, mn2 = inf;
    for(ll i = 0 ; i < a ;i ++)
    {cin >> arr[i] ; if(mn > arr[i])mn = arr[i];}
    for(ll i = 0; i < b ; i ++)
    {cin >> brr[i];if(mn2 > brr[i])mn2 = brr[i];}
    ans = mn + mn2;
    while(m --)
    {
        ll x ;
        cin >> a >> b >> x;
        ll t = arr[a - 1] + brr[b - 1] - x;
        if(t < ans)
            ans = t;
    }
    print(ans);
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ll t ;
    t = 1;
    //cin >> t;
    //seive();
    while(t --)
    {
        solve();
    }
    cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
    return 0;
}