#include<bits/stdc++.h>
#define N 2020
#define M 20200
using namespace std;

int n,to[M],hd[M],lk[N],ln[M],cst[M],u,v,tc,cnt=1,S,T,tp,q[N],h,t,pre[N];
long long dis[N],ans;
bool inq[N];

inline void add(int u,int v,int w,int c){to[++cnt]=v,hd[cnt]=lk[u],ln[cnt]=w,cst[cnt]=c,lk[u]=cnt;}

inline void Add(int u,int v,int w,int c){add(u,v,w,c),add(v,u,0,-c);}

int main()
{
	scanf("%d",&n);
	S=n*2,T=S+1;
	for(int i=0;i<S;i++)
	{
		scanf("%d%d%d",&u,&v,&tc);
		if(i<n)Add(S,i,tc,0);
		else Add(i,T+4,tc,0);
		for(int s=0;s<4;s++)
		{
			tp=(s&1?u:-u)+(s>>1&1?v:-v);
			if(i<n)Add(i,T+s,tc,tp);
			else Add(T+s,i,tc,-tp);
		}
	}T+=4;
	while(1)
	{
		for(int i=0;i<=T;i++)
		dis[i]=-1e17;
		dis[q[h=0]=S]=0,t=1;
		while(h^t)
		{
			inq[u=q[h++]]=0,h==N?h=0:0;
			for(int i=lk[u];i;i=hd[i])
				if(ln[i]&&dis[v=to[i]]<dis[u]+cst[i])
					dis[v]=dis[u]+cst[i],pre[v]=i,inq[v]?0:(inq[q[t++]=v]=1,t==N?t=0:0);
		}
		if(dis[T]==-1e17)break;
		tp=1e9;
		for(u=T;u^S;u=to[pre[u]^1])tp=min(tp,ln[pre[u]]);
		ans+=dis[T]*tp;
		for(u=pre[T];u;u=pre[to[u^1]])ln[u]-=tp,ln[u^1]+=tp;
	}
	printf("%lld",ans);
}

### Prompt

Generate 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>
#define N 2020
#define M 20200
using namespace std;

int n,to[M],hd[M],lk[N],ln[M],cst[M],u,v,tc,cnt=1,S,T,tp,q[N],h,t,pre[N];
long long dis[N],ans;
bool inq[N];

inline void add(int u,int v,int w,int c){to[++cnt]=v,hd[cnt]=lk[u],ln[cnt]=w,cst[cnt]=c,lk[u]=cnt;}

inline void Add(int u,int v,int w,int c){add(u,v,w,c),add(v,u,0,-c);}

int main()
{
	scanf("%d",&n);
	S=n*2,T=S+1;
	for(int i=0;i<S;i++)
	{
		scanf("%d%d%d",&u,&v,&tc);
		if(i<n)Add(S,i,tc,0);
		else Add(i,T+4,tc,0);
		for(int s=0;s<4;s++)
		{
			tp=(s&1?u:-u)+(s>>1&1?v:-v);
			if(i<n)Add(i,T+s,tc,tp);
			else Add(T+s,i,tc,-tp);
		}
	}T+=4;
	while(1)
	{
		for(int i=0;i<=T;i++)
		dis[i]=-1e17;
		dis[q[h=0]=S]=0,t=1;
		while(h^t)
		{
			inq[u=q[h++]]=0,h==N?h=0:0;
			for(int i=lk[u];i;i=hd[i])
				if(ln[i]&&dis[v=to[i]]<dis[u]+cst[i])
					dis[v]=dis[u]+cst[i],pre[v]=i,inq[v]?0:(inq[q[t++]=v]=1,t==N?t=0:0);
		}
		if(dis[T]==-1e17)break;
		tp=1e9;
		for(u=T;u^S;u=to[pre[u]^1])tp=min(tp,ln[pre[u]]);
		ans+=dis[T]*tp;
		for(u=pre[T];u;u=pre[to[u^1]])ln[u]-=tp,ln[u^1]+=tp;
	}
	printf("%lld",ans);
}

```