#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define REP(i,a,b) for(int i=(a),_end_=(b);i<=_end_;i++)
#define DREP(i,a,b) for(int i=(a),_end_=(b);i>=_end_;i--)
#define EREP(i,u) for(int i=start[u];i;i=e[i].next)
#define fi first
#define se second
#define mkr(a,b) make_pair(a,b)
#define SZ(A) ((int)A.size())
template<class T>inline void chkmin(T &a,T b){ if(a>b)a=b;}
template<class T>inline void chkmax(T &a,T b){ if(a<b)a=b;}
inline int read()
{
	int s=0,f=1;char ch=getchar();
	while(!isdigit(ch) && ch!='-')ch=getchar();
	if(ch=='-')ch=getchar(),f=-1;
	while(isdigit(ch))s=s*10+ch-'0',ch=getchar();
	return ~f?s:-s;
}

const int maxn=1e5+20;

struct node {
	int v,next,w;ll cost;
};
node e[maxn<<1];
int cnt=1,start[maxn];
inline void addedge(int u,int v,int w,ll cost){ e[++cnt]=(node){v,start[u],w,cost};start[u]=cnt;}
inline void Addedge(int u,int v,int w,ll cost){ addedge(u,v,w,cost); addedge(v,u,0,-cost);}

int n;
ll rx[maxn],ry[maxn],rc[maxn];
ll bx[maxn],by[maxn],bc[maxn];

int S,T,tot;
int Sn,Tn;
int MID[5],num1[maxn],num2[maxn];

//1:-,-
//2:+,+
//3:-,+
//4:+,-

inline void init()
{
	n=read();
	Sn=++tot; Tn=++tot;
	REP(i,1,4)MID[i]=++tot;
	REP(i,1,n)
	{
		rx[i]=read(); ry[i]=read(); rc[i]=read();
		num1[i]=++tot;
		Addedge(Sn,num1[i],rc[i],0);
		REP(j,1,4)
		{
			int res=0;
			if(j==1 || j==3)res-=rx[i]; else res+=rx[i];
			if(j==1 || j==4)res-=ry[i]; else res+=ry[i];
			res=-res;
			Addedge(num1[i],MID[j],rc[i],res);
		}
	}
	REP(i,1,n)
	{
		bx[i]=read(); by[i]=read(); bc[i]=read();
		num2[i]=++tot;
		Addedge(num2[i],Tn,bc[i],0);
		REP(j,1,4)
		{
			int res=0;
			if(j==1 || j==3)res-=bx[i]; else res+=bx[i];
			if(j==1 || j==4)res-=by[i]; else res+=by[i];
			Addedge(MID[j],num2[i],bc[i],res);
		}
	}
}

inline ll min_cost(int _S,int _T)
{
	ll ans=0;
	S=_S; T=_T;
	static ll dist[maxn];
	static int vis[maxn],pre[maxn];
	queue<int>q;
	while(1)
	{
		REP(i,1,tot)dist[i]=1ll<<60,vis[i]=0,pre[i]=0;
		q.push(S); dist[S]=0;
		while(!q.empty())
		{
			int u=q.front();q.pop();vis[u]=0;
			EREP(i,u)
			{
				int v=e[i].v;
				if(dist[v]>dist[u]+e[i].cost && e[i].w)
				{
					dist[v]=dist[u]+e[i].cost;
					pre[v]=i;
					if(!vis[v])vis[v]=1,q.push(v);
				}
			}
		}
		if(dist[T]==(1ll<<60))break;
		int flow=0x3f3f3f3f;
		for(int i=pre[T];i;i=pre[e[i^1].v])chkmin(flow,e[i].w);
		for(int i=pre[T];i;i=pre[e[i^1].v])e[i].w-=flow,e[i^1].w+=flow;
		ans+=1ll*dist[T]*flow;
	}
	return ans;
}

inline void doing()
{
	printf("%lld\n",-min_cost(Sn,Tn));
}

int main()
{
	init();
	doing();
	return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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;
#define ll long long
#define REP(i,a,b) for(int i=(a),_end_=(b);i<=_end_;i++)
#define DREP(i,a,b) for(int i=(a),_end_=(b);i>=_end_;i--)
#define EREP(i,u) for(int i=start[u];i;i=e[i].next)
#define fi first
#define se second
#define mkr(a,b) make_pair(a,b)
#define SZ(A) ((int)A.size())
template<class T>inline void chkmin(T &a,T b){ if(a>b)a=b;}
template<class T>inline void chkmax(T &a,T b){ if(a<b)a=b;}
inline int read()
{
	int s=0,f=1;char ch=getchar();
	while(!isdigit(ch) && ch!='-')ch=getchar();
	if(ch=='-')ch=getchar(),f=-1;
	while(isdigit(ch))s=s*10+ch-'0',ch=getchar();
	return ~f?s:-s;
}

const int maxn=1e5+20;

struct node {
	int v,next,w;ll cost;
};
node e[maxn<<1];
int cnt=1,start[maxn];
inline void addedge(int u,int v,int w,ll cost){ e[++cnt]=(node){v,start[u],w,cost};start[u]=cnt;}
inline void Addedge(int u,int v,int w,ll cost){ addedge(u,v,w,cost); addedge(v,u,0,-cost);}

int n;
ll rx[maxn],ry[maxn],rc[maxn];
ll bx[maxn],by[maxn],bc[maxn];

int S,T,tot;
int Sn,Tn;
int MID[5],num1[maxn],num2[maxn];

//1:-,-
//2:+,+
//3:-,+
//4:+,-

inline void init()
{
	n=read();
	Sn=++tot; Tn=++tot;
	REP(i,1,4)MID[i]=++tot;
	REP(i,1,n)
	{
		rx[i]=read(); ry[i]=read(); rc[i]=read();
		num1[i]=++tot;
		Addedge(Sn,num1[i],rc[i],0);
		REP(j,1,4)
		{
			int res=0;
			if(j==1 || j==3)res-=rx[i]; else res+=rx[i];
			if(j==1 || j==4)res-=ry[i]; else res+=ry[i];
			res=-res;
			Addedge(num1[i],MID[j],rc[i],res);
		}
	}
	REP(i,1,n)
	{
		bx[i]=read(); by[i]=read(); bc[i]=read();
		num2[i]=++tot;
		Addedge(num2[i],Tn,bc[i],0);
		REP(j,1,4)
		{
			int res=0;
			if(j==1 || j==3)res-=bx[i]; else res+=bx[i];
			if(j==1 || j==4)res-=by[i]; else res+=by[i];
			Addedge(MID[j],num2[i],bc[i],res);
		}
	}
}

inline ll min_cost(int _S,int _T)
{
	ll ans=0;
	S=_S; T=_T;
	static ll dist[maxn];
	static int vis[maxn],pre[maxn];
	queue<int>q;
	while(1)
	{
		REP(i,1,tot)dist[i]=1ll<<60,vis[i]=0,pre[i]=0;
		q.push(S); dist[S]=0;
		while(!q.empty())
		{
			int u=q.front();q.pop();vis[u]=0;
			EREP(i,u)
			{
				int v=e[i].v;
				if(dist[v]>dist[u]+e[i].cost && e[i].w)
				{
					dist[v]=dist[u]+e[i].cost;
					pre[v]=i;
					if(!vis[v])vis[v]=1,q.push(v);
				}
			}
		}
		if(dist[T]==(1ll<<60))break;
		int flow=0x3f3f3f3f;
		for(int i=pre[T];i;i=pre[e[i^1].v])chkmin(flow,e[i].w);
		for(int i=pre[T];i;i=pre[e[i^1].v])e[i].w-=flow,e[i^1].w+=flow;
		ans+=1ll*dist[T]*flow;
	}
	return ans;
}

inline void doing()
{
	printf("%lld\n",-min_cost(Sn,Tn));
}

int main()
{
	init();
	doing();
	return 0;
}

```