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++ (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define ll long long
#define int long long
using namespace std;
inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
	return x * f;
}
const int N = 1100;
int n,tot = 1,st,en;
queue<int> q;
struct node{int nex,to,val,flow;}a[N << 5];
int head[N << 2],flow[N << 2],dis[N << 2],vis[N << 2],pre[N << 2],last[N << 2];
int bfs()
{
	memset(flow,0x3f,sizeof(flow));
	for(int i = 0;i <= 2 * n + 5;i ++) dis[i] = -flow[i];
	pre[en] = -1; dis[st] = 0; q.push(st);
	while(!q.empty())
	{
		int f = q.front(); q.pop(); vis[f] = 0;
		for(int i = head[f],to;i;i = a[i].nex)
		{
			to = a[i].to;
			if(dis[to] < dis[f] + a[i].val && a[i].flow)
			{
				dis[to] = dis[f] + a[i].val; 
				flow[to] = min(flow[f],a[i].flow);
				pre[to] = f; last[to] = i;
				if(!vis[to]){vis[to] = 1; q.push(to);}
			}
		}
	}
	return pre[en] != -1;
}
void add(int f,int t,int v,int flow)
{
	a[++ tot].nex = head[f]; a[tot].to = t;
	a[tot].val = v; a[tot].flow = flow;
	head[f] = tot;
	a[++ tot].nex = head[t]; a[tot].to = f;
	a[tot].val = -v; a[tot].flow = 0;
	head[t] = tot;
}
signed main()
{
	n = read(); st = 0; en = 2 * n + 5; 
	for(int i = 1,x,y,c;i <= n;i ++)
	{
		x = read(); y = read(); c = read();
		add(st,i,0,c);
		add(i,2 * n + 1,x + y,c); add(i,2 * n + 2,x - y,c);
		add(i,2 * n + 3,-x + y,c);add(i,2 * n + 4,-x - y,c);
	}
	
	for(int i = 1,x,y,c;i <= n;i ++)
	{
		x = read(); y = read(); c = read();
		add(i + n,en,0,c);
		add(2 * n + 1,i + n,- x - y,c); add(2 * n + 2,i + n,-x + y,c);
		add(2 * n + 3,i + n,x - y,c); add(2 * n + 4,i + n,x + y,c);
	}
	int mdmf = 0,maxflow = 0;
	while(bfs())
	{
		mdmf += dis[en] * flow[en];
		maxflow += flow[en];
		int t = en;
		while(t != st)
		{
			a[last[t]].flow -= flow[en];
			a[last[t] ^ 1].flow += flow[en];
			t = pre[t];
		}
	}
	cout << mdmf << "\n";
	return 0;
}