#include<cstdio>
#include<cstring>
#include<iostream>
#include<vector>
using namespace std;
const int inf=1e9;
const long long INF=1e18;
const int dx[]={1,1,-1,-1};
const int dy[]={1,-1,1,-1};
int n,p,s,t; 
int head[2050];
struct edge
{
	int to,nex,cap,flow,cost;
	edge(int to=0,int nex=0,int cap=0,int flow=0,int cost=0):to(to),nex(nex),cap(cap),flow(flow),cost(cost){}
};
vector<edge> G;
inline void addedge(int u,int v,int c,int w)
{
	G.push_back(edge(v,head[u],c,0,w)),head[u]=G.size()-1;
	G.push_back(edge(u,head[v],0,0,-w)),head[v]=G.size()-1;
}
namespace MCMF
{
	int q[8050],hd,tl;
	int pre[2050];
	long long dis[2050];
	bool inq[2050];
	bool SPFA(int s,int t)
	{
		hd=tl=0;
		for(int i=0;i<p;++i)
		{
			dis[i]=-INF;
			pre[i]=-1;
			inq[i]=0;
		}
		dis[s]=0;
		inq[s]=1;
		q[tl++]=s;
		while(hd<tl)
		{
			int u=q[hd++];
			inq[u]=0;
			for(int i=head[u];~i;i=G[i].nex) if(G[i].cap!=G[i].flow)
			{
				int v=G[i].to;
				int w=G[i].cost;
				if(dis[v]<dis[u]+w)
				{
					dis[v]=dis[u]+w;
					pre[v]=i;
					if(!inq[v])
					{
						inq[v]=1;
						q[tl++]=v;
					}
				}
			}
		}
		return pre[t]!=-1;
	}
	void maxflow(int s,int t,int &flow,long long &cost)
	{
		flow=cost=0;
		while(SPFA(s,t))
		{
			int d=inf;
			for(int i=pre[t];~i;i=pre[G[i^1].to])
			{
  				d=min(d,G[i].cap-G[i].flow);
			}
			flow+=d;
			for(int i=pre[t];~i;i=pre[G[i^1].to])
			{
				G[i].flow+=d;
				G[i^1].flow-=d;
				cost+=1ll*d*G[i].cost;
			}
		}
	}
}
int main()
{
	scanf("%d",&n);
	p=4;
	s=p++;
	t=p++; 
	memset(head,-1,sizeof(head));
	for(int i=1;i<=n;++i)
	{
		int x,y,c; scanf("%d%d%d",&x,&y,&c);
		addedge(s,p,c,0);
		for(int k=0;k<4;++k)
		{
			addedge(p,k,inf,x*dx[k]+y*dy[k]);
		}
		++p;
	}
	for(int i=1;i<=n;++i)
	{
		int x,y,c; scanf("%d%d%d",&x,&y,&c);
		addedge(p,t,c,0);
		for(int k=0;k<4;++k)
		{
			addedge(k,p,inf,-x*dx[k]-y*dy[k]);
		}
		++p;
	}
	int flow;
	long long cost;
	MCMF::maxflow(s,t,flow,cost);
	printf("%lld\n",cost);
	return 0;
} 
### Prompt

Construct a cpp code solution to the problem outlined:

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<cstdio>
#include<cstring>
#include<iostream>
#include<vector>
using namespace std;
const int inf=1e9;
const long long INF=1e18;
const int dx[]={1,1,-1,-1};
const int dy[]={1,-1,1,-1};
int n,p,s,t; 
int head[2050];
struct edge
{
	int to,nex,cap,flow,cost;
	edge(int to=0,int nex=0,int cap=0,int flow=0,int cost=0):to(to),nex(nex),cap(cap),flow(flow),cost(cost){}
};
vector<edge> G;
inline void addedge(int u,int v,int c,int w)
{
	G.push_back(edge(v,head[u],c,0,w)),head[u]=G.size()-1;
	G.push_back(edge(u,head[v],0,0,-w)),head[v]=G.size()-1;
}
namespace MCMF
{
	int q[8050],hd,tl;
	int pre[2050];
	long long dis[2050];
	bool inq[2050];
	bool SPFA(int s,int t)
	{
		hd=tl=0;
		for(int i=0;i<p;++i)
		{
			dis[i]=-INF;
			pre[i]=-1;
			inq[i]=0;
		}
		dis[s]=0;
		inq[s]=1;
		q[tl++]=s;
		while(hd<tl)
		{
			int u=q[hd++];
			inq[u]=0;
			for(int i=head[u];~i;i=G[i].nex) if(G[i].cap!=G[i].flow)
			{
				int v=G[i].to;
				int w=G[i].cost;
				if(dis[v]<dis[u]+w)
				{
					dis[v]=dis[u]+w;
					pre[v]=i;
					if(!inq[v])
					{
						inq[v]=1;
						q[tl++]=v;
					}
				}
			}
		}
		return pre[t]!=-1;
	}
	void maxflow(int s,int t,int &flow,long long &cost)
	{
		flow=cost=0;
		while(SPFA(s,t))
		{
			int d=inf;
			for(int i=pre[t];~i;i=pre[G[i^1].to])
			{
  				d=min(d,G[i].cap-G[i].flow);
			}
			flow+=d;
			for(int i=pre[t];~i;i=pre[G[i^1].to])
			{
				G[i].flow+=d;
				G[i^1].flow-=d;
				cost+=1ll*d*G[i].cost;
			}
		}
	}
}
int main()
{
	scanf("%d",&n);
	p=4;
	s=p++;
	t=p++; 
	memset(head,-1,sizeof(head));
	for(int i=1;i<=n;++i)
	{
		int x,y,c; scanf("%d%d%d",&x,&y,&c);
		addedge(s,p,c,0);
		for(int k=0;k<4;++k)
		{
			addedge(p,k,inf,x*dx[k]+y*dy[k]);
		}
		++p;
	}
	for(int i=1;i<=n;++i)
	{
		int x,y,c; scanf("%d%d%d",&x,&y,&c);
		addedge(p,t,c,0);
		for(int k=0;k<4;++k)
		{
			addedge(k,p,inf,-x*dx[k]-y*dy[k]);
		}
		++p;
	}
	int flow;
	long long cost;
	MCMF::maxflow(s,t,flow,cost);
	printf("%lld\n",cost);
	return 0;
} 
```