Problem:
Score : 1200 points

Problem Statement

Snuke is playing with red and blue balls, placing them on a two-dimensional plane.

First, he performed N operations to place red balls. In the i-th of these operations, he placed RC_i red balls at coordinates (RX_i,RY_i).
Then, he performed another N operations to place blue balls. In the i-th of these operations, he placed BC_i blue balls at coordinates (BX_i,BY_i).
The total number of red balls placed and the total number of blue balls placed are equal, that is, \sum_{i=1}^{N} RC_i = \sum_{i=1}^{N} BC_i. Let this value be S.

Snuke will now form S pairs of red and blue balls so that every ball belongs to exactly one pair.
Let us define the score of a pair of a red ball at coordinates (rx, ry) and a blue ball at coordinates (bx, by) as |rx-bx| + |ry-by|.

Snuke wants to maximize the sum of the scores of the pairs.
Help him by finding the maximum possible sum of the scores of the pairs.

Constraints

1 \leq N \leq 1000

0 \leq RX_i,RY_i,BX_i,BY_i \leq 10^9

1 \leq RC_i,BC_i \leq 10

\sum_{i=1}^{N} RC_i = \sum_{i=1}^{N} BC_i

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
RX_1 RY_1 RC_1
RX_2 RY_2 RC_2
\vdots
RX_N RY_N RC_N
BX_1 BY_1 BC_1
BX_2 BY_2 BC_2
\vdots
BX_N BY_N BC_N

Output

Print the maximum possible sum of the scores of the pairs.

Sample Input 1

2
0 0 1
3 2 1
2 2 1
5 0 1

Sample Output 1

8

If we pair the red ball at coordinates (0,0) and the blue ball at coordinates (2,2), the score of this pair is |0-2| + |0-2|=4.
Then, if we pair the red ball at coordinates (3,2) and the blue ball at coordinates (5,0), the score of this pair is |3-5| + |2-0|=4.
Making these two pairs results in the total score of 8, which is the maximum result.

Sample Input 2

3
0 0 1
2 2 1
0 0 2
1 1 1
1 1 1
3 3 2

Sample Output 2

16

Snuke may have performed multiple operations at the same coordinates.

Sample Input 3

10
582463373 690528069 8
621230322 318051944 4
356524296 974059503 6
372751381 111542460 9
392867214 581476334 6
606955458 513028121 5
882201596 791660614 9
250465517 91918758 3
618624774 406956634 6
426294747 736401096 5
974896051 888765942 5
726682138 336960821 3
715144179 82444709 6
599055841 501257806 6
390484433 962747856 4
912334580 219343832 8
570458984 648862300 6
638017635 572157978 10
435958984 585073520 7
445612658 234265014 6

Sample Output 3

45152033546

Solution in Rust (1.15.1) (judge verdict: Accepted)

// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
#[allow(unused_macros)]
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)*}
    };
}

#[allow(unused_macros)]
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)*}
    };

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

#[allow(unused_macros)]
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>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

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

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

use std::cmp::min;
use std::collections::VecDeque;

const INF: isize = 1e12 as isize;


#[derive(Clone)]
struct FlowLink {
    src: usize,
    tgt: usize,
    cap: isize,
    cost: isize,
}

struct MinCostFlow {
    n: usize,
    links_from: Vec<Vec<usize>>,
    links: Vec<FlowLink>,
}

impl MinCostFlow {
    fn new(n: usize) -> MinCostFlow {
        MinCostFlow {
            n: n,
            links_from: vec![vec![]; n],
            links: vec![],
        }
    }
    fn add_link(&mut self, src: usize, tgt: usize, cap: isize, cost: isize) -> () {
        let i = self.links.len();
        self.links.push(FlowLink {
            src: src,
            tgt: tgt,
            cap: cap,
            cost: cost,
        });
        self.links.push(FlowLink {
            src: tgt,
            tgt: src,
            cap: 0,
            cost: -cost,
        });
        self.links_from[src].push(i);
        self.links_from[tgt].push(i + 1);
    }
    fn query(&mut self, s: usize, t: usize, quantity: isize) -> isize {
        let mut result: isize = 0;

        let mut remain = quantity;
        let mut predecessors = vec![0_usize; self.n];

        while remain > 0 {
//            println!("{}", remain);
            let mut distances: Vec<isize> = vec![INF; self.n];
            let mut updated: Vec<bool> = vec![false; self.n];
            distances[s] = 0;
            updated[s] = true;
            let mut deq: VecDeque<usize> = VecDeque::new();
            deq.push_back(s);

            while deq.len() > 0 {
                let v = deq.pop_front().unwrap();
                updated[v] = false;
                for li in self.links_from[v].iter() {
                    let link = &self.links[*li];
                    if link.cap == 0 { continue; }
                    let u = link.tgt;
                    let new_cost = distances[v] + link.cost;
                    if new_cost >= distances[u] { continue; }
                    distances[u] = new_cost;
                    predecessors[u] = *li;
                    if !updated[u] {
                        updated[u] = true;
                        deq.push_back(u);
                    }
                }
            }

            let mut min_cap = remain;
            let mut v = t;
            while v != s {
                let li = predecessors[v];
                min_cap = min(min_cap, self.links[li].cap);
                v = self.links[li].src;
            }
            v = t;
            while v != s {
                let li = predecessors[v];
                self.links[li].cap -= min_cap;
                self.links[li ^ 1].cap += min_cap;
                v = self.links[li].src;
            }
            remain -= min_cap;
            result -= min_cap * distances[t];
        };
        result
    }
}


fn main() {
    input! {
        n: usize,
        rrr: [(isize, isize, isize); n],
        bbb: [(isize, isize, isize); n],
    }

    let n2 = n * 2 + 6;
    let s = 0;
    let t = n * 2 + 1;
    let mut mcf = MinCostFlow::new(n2);
    let mut balls = 0;

    for (ri, (rx, ry, rc)) in (1..).zip(rrr) {
        balls += rc;
        mcf.add_link(s, ri, rc as isize, 0);
        mcf.add_link(ri, n2 - 4, rc, -(rx + ry));
        mcf.add_link(ri, n2 - 3, rc, -(-rx + ry));
        mcf.add_link(ri, n2 - 2, rc, -(rx - ry));
        mcf.add_link(ri, n2 - 1, rc, -(-rx - ry));
    }
    for (bi, (bx, by, bc)) in (n + 1..).zip(bbb) {
        mcf.add_link(bi, t, bc as isize, 0);
        mcf.add_link(n2 - 4, bi, bc, -(-bx - by));
        mcf.add_link(n2 - 3, bi, bc, -(bx - by));
        mcf.add_link(n2 - 2, bi, bc, -(-bx + by));
        mcf.add_link(n2 - 1, bi, bc, -(bx + by));
    }
    let res = mcf.query(s, t, balls);
    println!("{}", res);
}