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<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int S=0,T=2001,inf=0x3f3f3f3f;
const long long finf=0xc0c0c0c0c0c0c0c0ll;
struct bian
{
	int to,nxt,f,va;
}bi[3000030];
int n,head[2020],num=1,rx[1010],ry[1010],rc[1010],bx[1010],by[1010],bc[1010],from[2020];
long long dis[2020],ans;
bool bj[2020];
queue<int>q;
void add(int from,int to,int f,int va)
{
	bi[++num].nxt=head[from];
	head[from]=num;
	bi[num].to=to;
	bi[num].f=f;
	bi[num].va=va;
}
bool spfa()
{
	bj[S]=1;
	q.push(S);
	memset(dis,0xc0,sizeof(dis));
//	printf("%lld",dis[0]-1);
	dis[S]=0;
	int x,y;
	while(!q.empty())
	{
		x=q.front();q.pop();
		bj[x]=0;
		for(int i=head[x];i;i=bi[i].nxt)
			if(bi[i].f)
			{
				y=bi[i].to;
				if(dis[x]+bi[i].va>dis[y])
				{
					dis[y]=dis[x]+bi[i].va;
					from[y]=i;
					if(!bj[y])bj[y]=1,q.push(y);
				}
			}
	}
//	printf("%lld %lld %d\n",dis[T],0xc0c0c0c0c0c0c0c0ll,dis[T]>=0xc0c0c0c0c0c0c0c0ll);
	return dis[T]>finf;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
	{
		scanf("%d %d %d",&rx[i],&ry[i],&rc[i]);
		add(S,i,rc[i],0);
		add(i,S,0,0);
	} 
	for(int i=1;i<=n;++i)
	{
		scanf("%d %d %d",&bx[i],&by[i],&bc[i]);
		add(i+n,T,bc[i],0);
		add(T,i+n,0,0);
	}
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
			add(i,j+n,inf,abs(rx[i]-bx[j])+abs(ry[i]-by[j])),add(j+n,i,0,-abs(rx[i]-bx[j])-abs(ry[i]-by[j]));
	while(spfa())
	{
	//	printf("!");
		int x=T,flow=inf;
		while(x)flow=min(flow,bi[from[x]].f),x=bi[from[x]^1].to;
		ans+=1ll*flow*dis[T];
		x=T;
		while(x)bi[from[x]].f-=flow,bi[from[x]^1].f+=flow,x=bi[from[x]^1].to;
	}
	printf("%lld",ans);
	return 0;
}