#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
int n,p1,p2,p3,p4,S,T,tot=1;
const int N=1010,inf=1e9;
int head[N*2],to[N*20],nt[N*20],val[N*20],cost[N*20],pre[N*2],from[N*2],vis[N*2];
long long dis[N*2];
void ADD(int x,int y,int v,int c)
{
	to[++tot]=y;val[tot]=v;cost[tot]=c;nt[tot]=head[x];head[x]=tot;
}
void add(int x,int y,int v,int c)
{
	ADD(x,y,v,-c);ADD(y,x,0,c);
}
queue<int>q;
int SPFA()
{
	int x;
	memset(from,-1,sizeof(from));
	memset(dis,0x3f,sizeof(dis));
	while(!q.empty())q.pop();
	dis[S]=0;q.push(S);vis[S]=1;
	while(!q.empty())
	{
		x=q.front();q.pop();vis[x]=0;
		for(int i=head[x];i;i=nt[i])
			if(val[i]&&dis[to[i]]>dis[x]+cost[i])
			{
				dis[to[i]]=dis[x]+cost[i];from[to[i]]=x;pre[to[i]]=i;
				if(!vis[to[i]])q.push(to[i]),vis[to[i]]=1;
			}
	}
	return from[T]!=-1;
}
long long MFMC()
{
	long long ans=0;int k;
	while(SPFA())
	{
		k=inf;
		for(int i=T;i!=S;i=from[i])k=min(k,val[pre[i]]);
		ans+=dis[T]*k;
		for(int i=T;i!=S;i=from[i])val[pre[i]]-=k,val[pre[i]^1]+=k;
	}
	return ans;
}
int main()
{
	cin>>n;p1=2*n+1;p2=p1+1;p3=p1+2;p4=p1+3;S=p1+4;T=p1+5;
	for(int i=1,x,y,z;i<=n;++i)
	{
		scanf("%d%d%d",&x,&y,&z);
		add(S,i,z,0);
		add(i,p1,inf,x+y);
		add(i,p2,inf,x-y);
		add(i,p3,inf,-x+y);
		add(i,p4,inf,-x-y);
	}
	for(int i=1,x,y,z;i<=n;++i)
	{
		scanf("%d%d%d",&x,&y,&z);
		add(n+i,T,z,0);
		add(p1,n+i,inf,-x-y);
		add(p2,n+i,inf,-x+y);
		add(p3,n+i,inf,x-y);
		add(p4,n+i,inf,x+y);
	}
	cout<<-MFMC();
	return 0;
}
### Prompt

Please formulate 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<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
int n,p1,p2,p3,p4,S,T,tot=1;
const int N=1010,inf=1e9;
int head[N*2],to[N*20],nt[N*20],val[N*20],cost[N*20],pre[N*2],from[N*2],vis[N*2];
long long dis[N*2];
void ADD(int x,int y,int v,int c)
{
	to[++tot]=y;val[tot]=v;cost[tot]=c;nt[tot]=head[x];head[x]=tot;
}
void add(int x,int y,int v,int c)
{
	ADD(x,y,v,-c);ADD(y,x,0,c);
}
queue<int>q;
int SPFA()
{
	int x;
	memset(from,-1,sizeof(from));
	memset(dis,0x3f,sizeof(dis));
	while(!q.empty())q.pop();
	dis[S]=0;q.push(S);vis[S]=1;
	while(!q.empty())
	{
		x=q.front();q.pop();vis[x]=0;
		for(int i=head[x];i;i=nt[i])
			if(val[i]&&dis[to[i]]>dis[x]+cost[i])
			{
				dis[to[i]]=dis[x]+cost[i];from[to[i]]=x;pre[to[i]]=i;
				if(!vis[to[i]])q.push(to[i]),vis[to[i]]=1;
			}
	}
	return from[T]!=-1;
}
long long MFMC()
{
	long long ans=0;int k;
	while(SPFA())
	{
		k=inf;
		for(int i=T;i!=S;i=from[i])k=min(k,val[pre[i]]);
		ans+=dis[T]*k;
		for(int i=T;i!=S;i=from[i])val[pre[i]]-=k,val[pre[i]^1]+=k;
	}
	return ans;
}
int main()
{
	cin>>n;p1=2*n+1;p2=p1+1;p3=p1+2;p4=p1+3;S=p1+4;T=p1+5;
	for(int i=1,x,y,z;i<=n;++i)
	{
		scanf("%d%d%d",&x,&y,&z);
		add(S,i,z,0);
		add(i,p1,inf,x+y);
		add(i,p2,inf,x-y);
		add(i,p3,inf,-x+y);
		add(i,p4,inf,-x-y);
	}
	for(int i=1,x,y,z;i<=n;++i)
	{
		scanf("%d%d%d",&x,&y,&z);
		add(n+i,T,z,0);
		add(p1,n+i,inf,-x-y);
		add(p2,n+i,inf,-x+y);
		add(p3,n+i,inf,x-y);
		add(p4,n+i,inf,x+y);
	}
	cout<<-MFMC();
	return 0;
}
```