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 Rust (1.15.1) (judge verdict: Accepted)

#![allow(dead_code)]
#![allow(unused_macros)]

// 入力マクロ
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8 より
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

// #[allow(unused_imports)]
// use std::time::Instant;
// macro_rules! measure {
//     ( $x:expr) => {{
//         let start = Instant::now();
//         let result = $x;
//         let end = start.elapsed();
//         println!("Running in : {}.{:03}s", end.as_secs(), end.subsec_millis());
//         result
//     }};
// }

// ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
const MOD: u64 = 1_000_000_007;
fn main() {
    input! {
        a: usize, b: usize, m: usize,
        fr: [u64; a], mi: [u64; b],
        t: [(usize1, usize1, u64); m]
    }

    let ans = min(fr.iter().min().unwrap() + mi.iter().min().unwrap(), t.iter().map(|&(x, y, m)| fr[x] + mi[y] - m).min().unwrap());
    println!("{}", ans);
}


// ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

/// プログラムを終了
fn exit() {
    use std::process;
    process::exit(0)
}

/// 比較可能な2つの引数のうち大きい方を返す
fn max<T: PartialOrd>(n: T, m: T) -> T {
    if n >= m {
        n
    } else {
        m
    }
}

/// 比較[可能な2つの引数のうち小さい方を返す
fn min<T: PartialOrd>(n: T, m: T) -> T {
    if n <= m {
        n
    } else {
        m
    }
}

/// floor + as i64
fn into_floor(a: f64) -> i64 {
    (a - a % 1.0) as i64
}

/// ceil + as i64
fn into_ceil(a: f64) -> i64 {
    (a - a % 1.0) as i64 + 1i64
}

/// "Yes", "No" を出力
fn yes_no(flag: bool) {
    if flag {
        println!("Yes");
    } else {
        println!("No");
    }
}

/// 互除法を用いて最大公約数を求める
fn gcd(n: u64, m: u64) -> u64 {
    let (mut n, mut m) = (max(n, m), min(n, m));
    if m == 0 {
        return n;
    }
    let mut _r = 0;
    while n % m != 0 {
        _r = n % m;
        n = m;
        m = _r;
    }
    m
}

/// 最小公倍数を求める
fn lcm(n: u64, m: u64) -> u64 {
    n * m / gcd(n, m)
}

/// 配列全体の最大公約数
fn gcd_vec(v: &[u64]) -> u64 {
    let mut r = v[0];
    for i in v.iter().skip(1) {
        r = gcd(r, *i);
    }
    r
}

/// 配列全体の最小公倍数
fn lcm_vec(v: &[u64]) -> u64 {
    let mut r = v[0];
    for i in v.iter().skip(1) {
        r = lcm(r, *i);
    }
    r
}

/// Find the first factor (other than 1) of a number
fn firstfac(x: u64) -> u64 {
    if x % 2 == 0 {
        return 2;
    };
    // TODO: return to step_by
    // for n in (3..).step_by(2).take_while(|m| m*m <= x) {
    for n in (1..).map(|m| 2 * m + 1).take_while(|m| m * m <= x) {
        if x % n == 0 {
            return n;
        };
    }
    // No factor found. It must be prime.
    x
}

/// Test whether a number is prime. Checks every odd number up to `sqrt(n)`.
fn is_prime(n: u64) -> bool {
    if n <= 1 {
        return false;
    }
    firstfac(n) == n
}

// TODO!
// pub fn parmutation<T: PartialOrd>(v: &Vec<T>) -> Vec<Vec<T>> {
//     let mut sorted = v;

// }