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: Accepted)

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
#define PII pair<int,int>
#define MP make_pair
#define fir first
#define sec second
#define PB push_back
#define db long double
#define ll long long
using namespace std;
template <class T>
inline void rd(T &x) {
	x=0; char c=getchar(); int f=1;
	while(!isdigit(c)) { if(c=='-') f=-1; c=getchar(); }
	while(isdigit(c)) x=x*10-'0'+c,c=getchar(); x*=f;
}
const int N=1010,inf=1e9;
inline int Abs(int x) { return x>0?x:-x; }
int S,T;
namespace Flow {
	const int N=5010;
	int head[N],cur[N],vis[N];
	ll dis[N];
	struct ed { int to,next,f; ll w; };
	vector<ed> e;
	void init() { e.clear(); memset(head,-1,sizeof(head)); }
	void ad(int x,int y,int f,ll w) {
		e.PB((ed){y,head[x],f,w}); head[x]=e.size()-1;
		e.PB((ed){x,head[y],0,-w}); head[y]=e.size()-1;
	}
	queue<int> q;
	bool bfs() {
		memset(dis,-0x3f,sizeof(dis));
		memcpy(cur,head,sizeof(cur));
		dis[S]=0,q.push(S),vis[S]=1;
		while(!q.empty()) {
			int u=q.front(); q.pop(),vis[u]=0;
			for(int k=head[u];~k;k=e[k].next) if(e[k].f) {
				int v=e[k].to;
				if(dis[v]<dis[u]+e[k].w) {
					dis[v]=dis[u]+e[k].w;
					if(!vis[v]) q.push(v),vis[v]=1;
				}
			}
		}
		return dis[T]>-1e16;
	}
	int dfs(int u,int f) {
		if(u==T||!f) return f; int ret=0,tmp;
		vis[u]=1;
		for(int &k=cur[u];~k;k=e[k].next) if(e[k].f) {
			int v=e[k].to;
			if(dis[v]==dis[u]+e[k].w&&!vis[v]&&(tmp=dfs(v,min(f,e[k].f)))) {
				f-=tmp,ret+=tmp;
				e[k].f-=tmp,e[k^1].f+=tmp;
				if(!f) break;
			}
		}
		vis[u]=0;
		return ret;
	}
	ll work() { ll ans=0; while(bfs()) ans+=dis[T]*dfs(S,1e9); return ans; }
}
int rx[N],ry[N],rc[N];
int bx[N],by[N],bc[N];
int n;
int main() {
	rd(n);
	for(int i=1;i<=n;++i) rd(rx[i]),rd(ry[i]),rd(rc[i]);
	for(int i=1;i<=n;++i) rd(bx[i]),rd(by[i]),rd(bc[i]);
	S=2*n+1,T=2*n+2;
	int p[4];
	for(int i=0;i<4;++i) p[i]=2*n+3+i;
	Flow::init();
	for(int i=1;i<=n;++i) {
		Flow::ad(S,i,rc[i],0);
		Flow::ad(i+n,T,bc[i],0);
	}
	for(int i=1;i<=n;++i) {
		Flow::ad(i,p[0],inf,rx[i]+ry[i]);
		Flow::ad(i,p[1],inf,rx[i]-ry[i]);
		Flow::ad(i,p[2],inf,-rx[i]+ry[i]);
		Flow::ad(i,p[3],inf,-rx[i]-ry[i]);
		
		Flow::ad(p[0],i+n,inf,-bx[i]-by[i]);
		Flow::ad(p[1],i+n,inf,-bx[i]+by[i]);
		Flow::ad(p[2],i+n,inf,bx[i]-by[i]);
		Flow::ad(p[3],i+n,inf,bx[i]+by[i]);
	}
	printf("%lld",Flow::work());
	return 0;
}