#include<iostream>
#include<iomanip>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
int read()
{
	char c=getchar();
	int res=0;
	while(c<'0'||c>'9') c=getchar();
	while(c>='0'&&c<='9') res=(res<<1)+(res<<3)+(c^48),c=getchar();
	return res;
}
int n,h[3005],cnt=1,s=0,t,tot,A,B,C,D,pre[3005],p[3005];
long long d[3005],ans;
bool vis[3005];
struct Edge
{
	int to,next,d,v;
}w[200005];
struct que
{
	int x,y,c;
}a[100005],b[100005];
void add(int x,int y,int d,int v)
{
	w[++cnt]=(Edge){y,h[x],d,v};
	h[x]=cnt;
	w[++cnt]=(Edge){x,h[y],0,-v};
	h[y]=cnt;
}
bool Spfa()
{
	memset(d,-0x3f,sizeof(d));
	d[s]=0;p[s]=0x3f3f3f3f;
	queue<int> q;
	q.push(s);
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		vis[u]=0;
		for(int i=h[u];i;i=w[i].next)
		{
			int to=w[i].to;
			if(w[i].d&&d[to]<d[u]+w[i].v)
			{
				d[to]=d[u]+w[i].v;
				p[to]=min(p[u],w[i].d);pre[to]=i;
				if(!vis[to]) vis[to]=1,q.push(to);
			}
		}
	}
	return d[t]>0;
}
void Solve()
{
	ans+=d[t]*p[t];
	int now=t;
	while(now!=s)
	{
		w[pre[now]].d-=p[t];
		w[pre[now]^1].d+=p[t];
		now=w[pre[now]^1].to;
	}
}
void EK()
{
	while(Spfa()) Solve();
}
int main()
{
	n=read();
	for(int i=1;i<=n;i++) a[i].x=read(),a[i].y=read(),a[i].c=read();
	for(int i=1;i<=n;i++) b[i].x=read(),b[i].y=read(),b[i].c=read();
	tot=2*n;
	s=0,A=++tot,B=++tot,C=++tot,D=++tot;t=++tot;
	for(int i=1;i<=n;i++) add(s,i,a[i].c,0),add(i+n,t,b[i].c,0);
	for(int i=1;i<=n;i++)
	{
		int x=a[i].x,y=a[i].y;
		add(i,A,inf,x+y);
		add(i,B,inf,x-y);
		add(i,C,inf,-x+y);
		add(i,D,inf,-x-y);
	}
	for(int i=1;i<=n;i++)
	{
		int x=b[i].x,y=b[i].y;
		add(A,i+n,inf,-x-y);
		add(B,i+n,inf,-x+y);
		add(C,i+n,inf,x-y);
		add(D,i+n,inf,x+y);
	}
	EK();
	printf("%lld\n",ans);
}
### 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<iomanip>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
int read()
{
	char c=getchar();
	int res=0;
	while(c<'0'||c>'9') c=getchar();
	while(c>='0'&&c<='9') res=(res<<1)+(res<<3)+(c^48),c=getchar();
	return res;
}
int n,h[3005],cnt=1,s=0,t,tot,A,B,C,D,pre[3005],p[3005];
long long d[3005],ans;
bool vis[3005];
struct Edge
{
	int to,next,d,v;
}w[200005];
struct que
{
	int x,y,c;
}a[100005],b[100005];
void add(int x,int y,int d,int v)
{
	w[++cnt]=(Edge){y,h[x],d,v};
	h[x]=cnt;
	w[++cnt]=(Edge){x,h[y],0,-v};
	h[y]=cnt;
}
bool Spfa()
{
	memset(d,-0x3f,sizeof(d));
	d[s]=0;p[s]=0x3f3f3f3f;
	queue<int> q;
	q.push(s);
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		vis[u]=0;
		for(int i=h[u];i;i=w[i].next)
		{
			int to=w[i].to;
			if(w[i].d&&d[to]<d[u]+w[i].v)
			{
				d[to]=d[u]+w[i].v;
				p[to]=min(p[u],w[i].d);pre[to]=i;
				if(!vis[to]) vis[to]=1,q.push(to);
			}
		}
	}
	return d[t]>0;
}
void Solve()
{
	ans+=d[t]*p[t];
	int now=t;
	while(now!=s)
	{
		w[pre[now]].d-=p[t];
		w[pre[now]^1].d+=p[t];
		now=w[pre[now]^1].to;
	}
}
void EK()
{
	while(Spfa()) Solve();
}
int main()
{
	n=read();
	for(int i=1;i<=n;i++) a[i].x=read(),a[i].y=read(),a[i].c=read();
	for(int i=1;i<=n;i++) b[i].x=read(),b[i].y=read(),b[i].c=read();
	tot=2*n;
	s=0,A=++tot,B=++tot,C=++tot,D=++tot;t=++tot;
	for(int i=1;i<=n;i++) add(s,i,a[i].c,0),add(i+n,t,b[i].c,0);
	for(int i=1;i<=n;i++)
	{
		int x=a[i].x,y=a[i].y;
		add(i,A,inf,x+y);
		add(i,B,inf,x-y);
		add(i,C,inf,-x+y);
		add(i,D,inf,-x-y);
	}
	for(int i=1;i<=n;i++)
	{
		int x=b[i].x,y=b[i].y;
		add(A,i+n,inf,-x-y);
		add(B,i+n,inf,-x+y);
		add(C,i+n,inf,x-y);
		add(D,i+n,inf,x+y);
	}
	EK();
	printf("%lld\n",ans);
}
```