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)

/*
Author: QAQAutomaton
Lang: C++
Code: D.cpp
Mail: lk@qaq-am.com
Blog: https://www.qaq-am.com/
*/
#include<bits/stdc++.h>
#include<bits/extc++.h>
#define int long long
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef complex<double> cp;
typedef pair<int,int> pii;
int inf;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T,class T2>int chkmin(T &a,T2 b){return a>b?a=b,1:0;}
template<class T,class T2>int chkmax(T &a,T2 b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T,class T2>T mmin(T a,T2 b){return a<b?a:b;}
template<class T,class T2>T mmax(T a,T2 b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
struct __INIT__{
	__INIT__(){
		memset(&inf,0x3f,sizeof(inf));
	}
}__INIT___;
namespace io {
	const int SIZE = (1 << 21) + 1;
	char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
	// getchar
	#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
	// print the remaining part
	inline void flush () {
		fwrite (obuf, 1, oS - obuf, stdout);
		oS = obuf;
	}
	// putchar
	inline void putc (char x) {
		*oS ++ = x;
		if (oS == oT) flush ();
	}
	template<typename A>
	inline bool read (A &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
		return 1;
	}
	inline bool read (char &x) {
		while((x=gc())==' '||x=='\n' || x=='\r');
		return x!=EOF;
	}
	inline bool read(char *x){
		while((*x=gc())=='\n' || *x==' '||*x=='\r');
		if(*x==EOF)return 0;
		while(!(*x=='\n'||*x==' '||*x=='\r'||*x==EOF))*(++x)=gc();
		*x=0;
		return 1;
	}
	template<typename A,typename ...B>
	inline bool read(A &x,B &...y){
		return read(x)&&read(y...);
	}
	template<typename A>
	inline bool write (A x) {
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
		while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
		while (qr) putc (qu[qr --]);
		return 0;
	}
	inline bool write (char x) {
		putc(x);
		return 0;
	}
	inline bool write(const char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	inline bool write(char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	template<typename A,typename ...B>
	inline bool write(A x,B ...y){
		return write(x)||write(y...);
	}
	//no need to call flush at the end manually!
	struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
namespace mcmf{
	int n;
	int to[100005],lst[100005],beg[2005],w[100005],c[100005],e=1;
	void add(int s,int t,int fl,int co){
		to[++e]=t;lst[e]=beg[s];beg[s]=e;w[e]=fl;c[e]=co;
		to[++e]=s;lst[e]=beg[t];beg[t]=e;w[e]=0;c[e]=-co;
	}
	int cur[2005],d[2005],inq[2005];
	int q[2005],*las=q+2004,*l,*r;
	int fl[2005],fr[2005];
	int spfa(int s,int t){
		for(int i=1;i<=n;++i)d[i]=inf;
		d[s]=0;
		*(l=r=q)=s;
		fl[s]=inf;
		for(;;l==las?l=q:++l){
			inq[*l]=0;
			for(int i=beg[*l];i;i=lst[i])if(w[i]){
				if(chkmin(d[to[i]],d[*l]+c[i])){
					if(!inq[to[i]]){
						*(r==las?r=q:++r)=to[i];

						inq[to[i]]=1;
					}
					fr[to[i]]=i;
					fl[to[i]]=min(fl[*l],w[i]);
				}
			}
			if(l==r)break;
		}
		return d[t];
	}
	__gnu_pbds::priority_queue<pii,greater<pii> >pq;
	__gnu_pbds::priority_queue<pii,greater<pii> >::point_iterator it[2005];
	int dijk(int s,int t){
		for(int i=1;i<=n;++i){cur[i]=beg[i];d[i]=inf;}
		d[s]=0;
		pq.clear();
		fl[s]=inf;
		for(int i=1;i<=n;++i)it[i]=pq.push(make_pair(d[i],i));
		while(!pq.empty()){
			if(pq.top().x==inf)return inf;
			int x=pq.top().y;
			if(x==t)return pq.top().x;
			pq.pop();
			for(int i=beg[x];i;i=lst[i])if(w[i]&&chkmin(d[to[i]],d[x]+c[i])){
				assert(c[i]>=0);
				pq.modify(it[to[i]],make_pair(d[to[i]],to[i]));
				fr[to[i]]=i;
				fl[to[i]]=min(fl[x],w[i]);
			}
		}
		return d[t];
	}
	void update(){
		for(int i=2;i<=e;++i){
			c[i]+=d[to[i^1]]-d[to[i]];	
		}	
	}
	int dfs(int s,int t,int fl){
		if(s==t)return fl;
		for(int &i=cur[s];i;i=lst[i])if(w[i] && d[s]+c[i]==d[to[i]]){
			int f=dfs(to[i],t,min(fl,w[i]));
			if(f){w[i]-=f;w[i^1]+=f;return f;}
		}
		return 0;
	}
	int calc(int s,int t){
		int cur=spfa(s,t),nex,ans=0;
		while((nex=spfa(s,t))!=inf){
			ans+=nex*fl[t];
			for(int i=t;i!=s;i=to[fr[i]^1]){
				w[fr[i]]-=fl[t];
				w[fr[i]^1]+=fl[t];
			}
		}		
		return ans;
	}
}
signed main(){
#ifdef QAQAutoMaton 
	freopen("D.in","r",stdin);
	freopen("D.out","w",stdout);
#endif
	int n,x,y,c;
	read(n);
	int m=6;
	for(int i=1;i<=n;++i){
		read(x,y,c);
		x=-x;y=-y;
		++m;
		mcmf::add(1,m,c,0);
		mcmf::add(m,3,inf,x+y);
		mcmf::add(m,4,inf,x-y);
		mcmf::add(m,5,inf,-x+y);
		mcmf::add(m,6,inf,-x-y);
	}
	for(int i=1;i<=n;++i){
		read(x,y,c);
		++m;
		mcmf::add(m,2,c,0);
		mcmf::add(3,m,inf,x+y);
		mcmf::add(4,m,inf,x-y);
		mcmf::add(5,m,inf,-x+y);
		mcmf::add(6,m,inf,-x-y);
	}
	mcmf::n=m;
	write(-mcmf::calc(1,2),'\n');
	return 0;
}