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 C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int dir[4][2] = {1, -1, -1, 1, -1, -1, 1, 1};
const int kN = 10010, kInf = 1e18, kNode = 1e5;
const int inf = kInf;
int node_cnt, n;
int id1[4], id2[4];
int x[kN], y[kN], w[kN];

struct Edge {
	int to, cap, flow, cost, rev;
	Edge(int to, int cap, int cost, int rev):
		to(to), cap(cap), flow(0), cost(cost), rev(rev) {}
};
vector<Edge> graph[kNode];
int dis[kNode], st, ed;
bool in_que[kNode], in_stk[kNode];
queue<int> que;

void AddEdge(int from, int to, int cap, int cost) {
	graph[from].push_back(Edge(to, cap, cost, graph[to].size()));
	graph[to].push_back(Edge(from, 0, -cost, graph[from].size() - 1));
}

bool Spfa() {
	for (int i = 1; i <= node_cnt; ++i) dis[i] = i == st ? 0 : kInf;
	que.push(st);
	while (!que.empty()) {
		int p = que.front();
		que.pop();
		in_que[p] = 0;
		for (int i = 0; i < (int)graph[p].size(); ++i) {
			Edge e = graph[p][i];
			if (e.cap > e.flow && dis[p] + e.cost < dis[e.to]) {
				dis[e.to] = dis[p] + e.cost;
				if (!in_que[e.to]) {
					in_que[e.to] = true;
					que.push(e.to);
				}
			}
		}
	}
	return dis[ed] < kInf;
}

int Dfs(int p, int a) {
	if (p == ed) return a;
	in_stk[p] = true;
	int ans = 0, now;
	for (int i = 0; i < (int)graph[p].size(); ++i) {
		Edge &e = graph[p][i];
		if (e.cap > e.flow && !in_stk[e.to] &&
				dis[p] + e.cost == dis[e.to]) {
			now = Dfs(e.to, min(a, e.cap - e.flow));
			ans += now;
			a -= now;
			e.flow += now;
			graph[e.to][e.rev].flow -= now;
			if (!a) break;
		}
	}
	in_stk[p] = false;
	return ans;
}

int MinCostMaxFlow() {
	int ans = 0;
	while (Spfa())
		ans += dis[ed] * Dfs(st, kInf);
	return ans;
}

signed main() {
	st = ++node_cnt;
	ed = ++node_cnt;
	scanf("%lld", &n);
	for (int i = 0; i < 4; ++i) {
		id1[i] = ++node_cnt;
		id2[i] = ++node_cnt;
	}
	for (int i = 1; i <= n * 2; ++i) {
		scanf("%lld%lld%lld", &x[i], &y[i], &w[i]);
		++node_cnt;
		if (i <= n) AddEdge(st, node_cnt, w[i], 0);
		else AddEdge(node_cnt, ed, w[i], 0);
		for (int j = 0; j < 4; ++j) {
			int w = x[i] * dir[j][0] + y[i] * dir[j][1];
			if (i <= n) AddEdge(node_cnt, id1[j], inf, -w);
			else AddEdge(id2[j], node_cnt, inf, -w);
		}
	}
	for (int i = 0; i < 4; ++i)
		AddEdge(id1[i], id2[i ^ 1], inf, 0);
	printf("%lld", -MinCostMaxFlow());
}