#include <bits/stdc++.h>

using namespace std;
#define gc getchar()
#define re register
#define int long long
inline int read()
{
	re int x(0);re char c(gc);
	while(c>'9'||c<'0')c=gc;
	while(c<='9'&&c>='0')x=x*10+c-48,c=gc;
	return x;
}

const int N=1e5+10;

namespace maxflow_cost
{
	const int INF=1e15+10;
	int n,h[N],cnt=1,s,t,dis[N],vis[N],pre[N],f[N];
	struct edge{int next,to,f,c;}e[N<<1];
	#define QXX(u) for(int i=h[u],v;v=e[i].to,i;i=e[i].next)
	void add(int u,int v,int f,int c)
	{
		e[++cnt]=(edge){h[u],v,f, c};h[u]=cnt;
		e[++cnt]=(edge){h[v],u,0,-c},h[v]=cnt;
	}
	queue <int> q;
	int SPFA()
	{
		queue<int> q;
		while(!q.empty()) q.pop();
		q.push(s);
		for(int i=1;i<=t;++i)
			dis[i]=INF,vis[i]=0;
		vis[s]=1,dis[s]=0,pre[t]=-1,f[s]=INF;
		while(!q.empty())
		{
			int u=q.front();
			q.pop(); vis[u]=0;
			QXX(u) if(e[i].f) {
				if(dis[v]>dis[u]+e[i].c) {
					dis[v]=dis[u]+e[i].c;
					f[v]=min(f[u],e[i].f);
					pre[v]=i;
					if(!vis[v])
						vis[v]=1,q.push(v);
				}
			}
		}
	    return pre[t]==-1?-1:f[t];
	}
	int dinic()
	{
		int in,maxf=0,maxc=0,x,y;
		while((in=SPFA())!=-1) {
			x=t;
			while(x!=s) {
				y=pre[x];
				e[y].f-=in;
				e[y^1].f+=in;
				x=e[y^1].to;
			}
			maxf+=in;
			maxc+=in*dis[t];
		}
		return -maxc;
	}
}

using namespace maxflow_cost;

signed main()
{
	n=read();
	s=n*2+5,t=s+1;
	for(int i=1;i<=n;++i)
	{
		int x=read(),y=read(),c=read();
		add(s,i,c,0);
		add(i,2*n+1,c,+x+y);
		add(i,2*n+2,c,+x-y);
		add(i,2*n+3,c,-x+y);
		add(i,2*n+4,c,-x-y);
	}
	for(int i=1;i<=n;++i)
	{
		int x=read(),y=read(),c=read(),id=i+n;
		add(id,t,c,0);
		add(2*n+1,id,c,-x-y);
		add(2*n+2,id,c,-x+y);
		add(2*n+3,id,c,+x-y);
		add(2*n+4,id,c,+x+y);
	}
	cout<<dinic()<<'\n';
	return 0;
}
### 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>

using namespace std;
#define gc getchar()
#define re register
#define int long long
inline int read()
{
	re int x(0);re char c(gc);
	while(c>'9'||c<'0')c=gc;
	while(c<='9'&&c>='0')x=x*10+c-48,c=gc;
	return x;
}

const int N=1e5+10;

namespace maxflow_cost
{
	const int INF=1e15+10;
	int n,h[N],cnt=1,s,t,dis[N],vis[N],pre[N],f[N];
	struct edge{int next,to,f,c;}e[N<<1];
	#define QXX(u) for(int i=h[u],v;v=e[i].to,i;i=e[i].next)
	void add(int u,int v,int f,int c)
	{
		e[++cnt]=(edge){h[u],v,f, c};h[u]=cnt;
		e[++cnt]=(edge){h[v],u,0,-c},h[v]=cnt;
	}
	queue <int> q;
	int SPFA()
	{
		queue<int> q;
		while(!q.empty()) q.pop();
		q.push(s);
		for(int i=1;i<=t;++i)
			dis[i]=INF,vis[i]=0;
		vis[s]=1,dis[s]=0,pre[t]=-1,f[s]=INF;
		while(!q.empty())
		{
			int u=q.front();
			q.pop(); vis[u]=0;
			QXX(u) if(e[i].f) {
				if(dis[v]>dis[u]+e[i].c) {
					dis[v]=dis[u]+e[i].c;
					f[v]=min(f[u],e[i].f);
					pre[v]=i;
					if(!vis[v])
						vis[v]=1,q.push(v);
				}
			}
		}
	    return pre[t]==-1?-1:f[t];
	}
	int dinic()
	{
		int in,maxf=0,maxc=0,x,y;
		while((in=SPFA())!=-1) {
			x=t;
			while(x!=s) {
				y=pre[x];
				e[y].f-=in;
				e[y^1].f+=in;
				x=e[y^1].to;
			}
			maxf+=in;
			maxc+=in*dis[t];
		}
		return -maxc;
	}
}

using namespace maxflow_cost;

signed main()
{
	n=read();
	s=n*2+5,t=s+1;
	for(int i=1;i<=n;++i)
	{
		int x=read(),y=read(),c=read();
		add(s,i,c,0);
		add(i,2*n+1,c,+x+y);
		add(i,2*n+2,c,+x-y);
		add(i,2*n+3,c,-x+y);
		add(i,2*n+4,c,-x-y);
	}
	for(int i=1;i<=n;++i)
	{
		int x=read(),y=read(),c=read(),id=i+n;
		add(id,t,c,0);
		add(2*n+1,id,c,-x-y);
		add(2*n+2,id,c,-x+y);
		add(2*n+3,id,c,+x-y);
		add(2*n+4,id,c,+x+y);
	}
	cout<<dinic()<<'\n';
	return 0;
}
```