#include<bits/stdc++.h>
using namespace std;

typedef long long lint;
typedef long double louble;

template<typename T1,typename T2> inline T1 max(T1 a,T2 b){return a<b?b:a;}
template<typename T1,typename T2> inline T1 min(T1 a,T2 b){return a<b?a:b;}

namespace ae86
{
	const int bufl = 1<<15;
	
	char buf[bufl],*s=buf,*t=buf;
	
	inline int fetch()
	{
		if(s==t){t=(s=buf)+fread(buf,1,bufl,stdin);if(s==t)return EOF;}
		return *s++;
	}
	
	inline int ty()
	{
		int a=0,b=1,c=fetch();
		while(!isdigit(c))b^=c=='-',c=fetch();
		while(isdigit(c))a=a*10+c-48,c=fetch();
		return b?a:-a;
	}
}
using ae86::ty;

const int _ = 1007;
const lint linf = 0x3f3f3f3f3f3f3f3f;

struct costflow
{
	static const int _n = _ + _ , _e = 999999;

	int to[_e],ne[_e],he[_n],ecnt;
	lint ev[_e],ec[_e];

	inline void adde(int a,int b,lint c,lint d)
	{
		to[++ecnt]=b,ev[ecnt]=c,ec[ecnt]=d,ne[ecnt]=he[a],he[a]=ecnt;
	}

	inline void addde(int a,int b,lint c,lint d)
	{
		adde(a,b,c,d),adde(b,a,0,-d);
	}

	int dcnt,S,T,pv[_n],pe[_n];
	lint dis[_n];bool ed[_n];

	inline int spfa()
	{
		memset(dis,63,sizeof(lint)*(dcnt+2));
		memset(ed,0,sizeof(bool)*(dcnt+2));

		queue<int> q;
		q.emplace(S),dis[S]=0,ed[S]=1;
		while(!q.empty())
		{
			int a=q.front();q.pop(),ed[a]=0;
			for(int i=he[a];i;i=ne[i])
			{
				if(!ev[i])continue;
				int b=to[i];
				if(dis[b]<=dis[a]+ec[i])continue;
				dis[b]=dis[a]+ec[i],pe[b]=i,pv[b]=a;
				if(!ed[b])q.emplace(b),ed[b]=1;
			}
		}

		return dis[T]<linf;
	}

	inline pair<lint,lint> run(int s,int t)
	{
		S=s,T=t;
		lint sflw=0,scos=0;
		while(spfa())
		{
			lint dlt=linf;
			for(int i=T;i!=S;i=pv[i])dlt=min(dlt,ev[pe[i]]);
			for(int i=T;i!=S;i=pv[i])scos+=dlt*ec[pe[i]],ev[pe[i]]-=dlt,ev[pe[i]^1]+=dlt;
			sflw+=dlt;
		}
		return make_pair(sflw,scos);
	}
	
	costflow()
	{
		memset(he,0,sizeof(he)),ecnt=1;
	}
}f;

int n,nn,xs[_],ys[_],cs[_],xt[_],yt[_],ct[_];

int main()
{
	n=ty(),nn=n+n;
	for(int i=1;i<=n;i++)xs[i]=ty(),ys[i]=ty(),cs[i]=ty();
	for(int i=1;i<=n;i++)xt[i]=ty(),yt[i]=ty(),ct[i]=ty();

	f.dcnt=nn+4;

	int s=++f.dcnt,t=++f.dcnt;

	for(int i=1;i<=n;i++)
	{
		f.addde(s,i,cs[i],0);
		f.addde(i,nn+1,cs[i],-xs[i]-ys[i]);
		f.addde(i,nn+2,cs[i],+xs[i]-ys[i]);
		f.addde(i,nn+3,cs[i],-xs[i]+ys[i]);
		f.addde(i,nn+4,cs[i],+xs[i]+ys[i]);
	}

	for(int i=1;i<=n;i++)
	{
		f.addde(n+i,t,ct[i],0);
		f.addde(nn+1,i+n,ct[i],+xt[i]+yt[i]);
		f.addde(nn+2,i+n,ct[i],-xt[i]+yt[i]);
		f.addde(nn+3,i+n,ct[i],+xt[i]-yt[i]);
		f.addde(nn+4,i+n,ct[i],-xt[i]-yt[i]);
	}

	auto ans=f.run(s,t);
	printf("%lld\n",-ans.second);

	return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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.

Examples

Input

2
0 0 1
3 2 1
2 2 1
5 0 1


Output

8


Input

3
0 0 1
2 2 1
0 0 2
1 1 1
1 1 1
3 3 2


Output

16


Input

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


Output

45152033546

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

typedef long long lint;
typedef long double louble;

template<typename T1,typename T2> inline T1 max(T1 a,T2 b){return a<b?b:a;}
template<typename T1,typename T2> inline T1 min(T1 a,T2 b){return a<b?a:b;}

namespace ae86
{
	const int bufl = 1<<15;
	
	char buf[bufl],*s=buf,*t=buf;
	
	inline int fetch()
	{
		if(s==t){t=(s=buf)+fread(buf,1,bufl,stdin);if(s==t)return EOF;}
		return *s++;
	}
	
	inline int ty()
	{
		int a=0,b=1,c=fetch();
		while(!isdigit(c))b^=c=='-',c=fetch();
		while(isdigit(c))a=a*10+c-48,c=fetch();
		return b?a:-a;
	}
}
using ae86::ty;

const int _ = 1007;
const lint linf = 0x3f3f3f3f3f3f3f3f;

struct costflow
{
	static const int _n = _ + _ , _e = 999999;

	int to[_e],ne[_e],he[_n],ecnt;
	lint ev[_e],ec[_e];

	inline void adde(int a,int b,lint c,lint d)
	{
		to[++ecnt]=b,ev[ecnt]=c,ec[ecnt]=d,ne[ecnt]=he[a],he[a]=ecnt;
	}

	inline void addde(int a,int b,lint c,lint d)
	{
		adde(a,b,c,d),adde(b,a,0,-d);
	}

	int dcnt,S,T,pv[_n],pe[_n];
	lint dis[_n];bool ed[_n];

	inline int spfa()
	{
		memset(dis,63,sizeof(lint)*(dcnt+2));
		memset(ed,0,sizeof(bool)*(dcnt+2));

		queue<int> q;
		q.emplace(S),dis[S]=0,ed[S]=1;
		while(!q.empty())
		{
			int a=q.front();q.pop(),ed[a]=0;
			for(int i=he[a];i;i=ne[i])
			{
				if(!ev[i])continue;
				int b=to[i];
				if(dis[b]<=dis[a]+ec[i])continue;
				dis[b]=dis[a]+ec[i],pe[b]=i,pv[b]=a;
				if(!ed[b])q.emplace(b),ed[b]=1;
			}
		}

		return dis[T]<linf;
	}

	inline pair<lint,lint> run(int s,int t)
	{
		S=s,T=t;
		lint sflw=0,scos=0;
		while(spfa())
		{
			lint dlt=linf;
			for(int i=T;i!=S;i=pv[i])dlt=min(dlt,ev[pe[i]]);
			for(int i=T;i!=S;i=pv[i])scos+=dlt*ec[pe[i]],ev[pe[i]]-=dlt,ev[pe[i]^1]+=dlt;
			sflw+=dlt;
		}
		return make_pair(sflw,scos);
	}
	
	costflow()
	{
		memset(he,0,sizeof(he)),ecnt=1;
	}
}f;

int n,nn,xs[_],ys[_],cs[_],xt[_],yt[_],ct[_];

int main()
{
	n=ty(),nn=n+n;
	for(int i=1;i<=n;i++)xs[i]=ty(),ys[i]=ty(),cs[i]=ty();
	for(int i=1;i<=n;i++)xt[i]=ty(),yt[i]=ty(),ct[i]=ty();

	f.dcnt=nn+4;

	int s=++f.dcnt,t=++f.dcnt;

	for(int i=1;i<=n;i++)
	{
		f.addde(s,i,cs[i],0);
		f.addde(i,nn+1,cs[i],-xs[i]-ys[i]);
		f.addde(i,nn+2,cs[i],+xs[i]-ys[i]);
		f.addde(i,nn+3,cs[i],-xs[i]+ys[i]);
		f.addde(i,nn+4,cs[i],+xs[i]+ys[i]);
	}

	for(int i=1;i<=n;i++)
	{
		f.addde(n+i,t,ct[i],0);
		f.addde(nn+1,i+n,ct[i],+xt[i]+yt[i]);
		f.addde(nn+2,i+n,ct[i],-xt[i]+yt[i]);
		f.addde(nn+3,i+n,ct[i],+xt[i]-yt[i]);
		f.addde(nn+4,i+n,ct[i],-xt[i]-yt[i]);
	}

	auto ans=f.run(s,t);
	printf("%lld\n",-ans.second);

	return 0;
}

```