#include<bits/stdc++.h>
using namespace std;

const long long inf=1000000000000000000ll;

namespace flow
{
	struct node{int to,next,c; long long val;}e[200010];
	int n,cnt,hd[3010],q[3000010],l,r,x,la[3010];
	long long dis[3010];
	bool inq[3010];
	void init(int N)
	{
		n=N,cnt=1;
		for (int i=1; i<=n; i++) hd[i]=0;
	}
	void addedge(int x,int y,int c,long long v)
	{
		e[++cnt]=(node){y,hd[x],c,v},hd[x]=cnt;
		e[++cnt]=(node){x,hd[y],0,-v},hd[y]=cnt;
	}
	pair<int,long long> solve()
	{
		int f=0;
		long long ans=0;
		while (1)
		{
			dis[1]=0,q[l=r=1]=1,inq[1]=1,la[1]=0;
			for (int i=2; i<=n; i++) inq[i]=0,dis[i]=inf,la[i]=0;
			while (l<=r)
			{
				x=q[l++];
				for (int i=hd[x]; i; i=e[i].next)
					if (e[i].c&&dis[e[i].to]>dis[x]+e[i].val)
					{
						dis[e[i].to]=dis[x]+e[i].val,la[e[i].to]=i;
						if (!inq[e[i].to]) inq[e[i].to]=1,q[++r]=e[i].to;
					}
				inq[x]=0;
			}
			if (dis[n]==inf) break;
			x=n; int nw=1000000000;
			while (la[x]) nw=min(nw,e[la[x]].c),x=e[la[x]^1].to;
			x=n,f+=nw;
			while (la[x]) e[la[x]].c-=nw,e[la[x]^1].c+=nw,ans+=e[la[x]].val*nw,x=e[la[x]^1].to;
		}
		return make_pair(f,ans);
	}
}

int n,x,y,c;

int main()
{
	scanf("%d",&n);
	flow::init(2*n+6);
	for (int i=1; i<=n; i++) 
	{
		scanf("%d%d%d",&x,&y,&c);
		flow::addedge(1,i+1,c,0);
		flow::addedge(i+1,2*n+2,c,x+y);
		flow::addedge(i+1,2*n+3,c,x-y);
		flow::addedge(i+1,2*n+4,c,-x+y);
		flow::addedge(i+1,2*n+5,c,-x-y);
	}
	for (int i=1; i<=n; i++)
	{
		scanf("%d%d%d",&x,&y,&c);
		flow::addedge(i+n+1,2*n+6,c,0);
		flow::addedge(2*n+2,i+n+1,c,-x-y);
		flow::addedge(2*n+3,i+n+1,c,-x+y);
		flow::addedge(2*n+4,i+n+1,c,x-y);
		flow::addedge(2*n+5,i+n+1,c,x+y);		
	}
	printf("%lld\n",-flow::solve().second);
	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;

const long long inf=1000000000000000000ll;

namespace flow
{
	struct node{int to,next,c; long long val;}e[200010];
	int n,cnt,hd[3010],q[3000010],l,r,x,la[3010];
	long long dis[3010];
	bool inq[3010];
	void init(int N)
	{
		n=N,cnt=1;
		for (int i=1; i<=n; i++) hd[i]=0;
	}
	void addedge(int x,int y,int c,long long v)
	{
		e[++cnt]=(node){y,hd[x],c,v},hd[x]=cnt;
		e[++cnt]=(node){x,hd[y],0,-v},hd[y]=cnt;
	}
	pair<int,long long> solve()
	{
		int f=0;
		long long ans=0;
		while (1)
		{
			dis[1]=0,q[l=r=1]=1,inq[1]=1,la[1]=0;
			for (int i=2; i<=n; i++) inq[i]=0,dis[i]=inf,la[i]=0;
			while (l<=r)
			{
				x=q[l++];
				for (int i=hd[x]; i; i=e[i].next)
					if (e[i].c&&dis[e[i].to]>dis[x]+e[i].val)
					{
						dis[e[i].to]=dis[x]+e[i].val,la[e[i].to]=i;
						if (!inq[e[i].to]) inq[e[i].to]=1,q[++r]=e[i].to;
					}
				inq[x]=0;
			}
			if (dis[n]==inf) break;
			x=n; int nw=1000000000;
			while (la[x]) nw=min(nw,e[la[x]].c),x=e[la[x]^1].to;
			x=n,f+=nw;
			while (la[x]) e[la[x]].c-=nw,e[la[x]^1].c+=nw,ans+=e[la[x]].val*nw,x=e[la[x]^1].to;
		}
		return make_pair(f,ans);
	}
}

int n,x,y,c;

int main()
{
	scanf("%d",&n);
	flow::init(2*n+6);
	for (int i=1; i<=n; i++) 
	{
		scanf("%d%d%d",&x,&y,&c);
		flow::addedge(1,i+1,c,0);
		flow::addedge(i+1,2*n+2,c,x+y);
		flow::addedge(i+1,2*n+3,c,x-y);
		flow::addedge(i+1,2*n+4,c,-x+y);
		flow::addedge(i+1,2*n+5,c,-x-y);
	}
	for (int i=1; i<=n; i++)
	{
		scanf("%d%d%d",&x,&y,&c);
		flow::addedge(i+n+1,2*n+6,c,0);
		flow::addedge(2*n+2,i+n+1,c,-x-y);
		flow::addedge(2*n+3,i+n+1,c,-x+y);
		flow::addedge(2*n+4,i+n+1,c,x-y);
		flow::addedge(2*n+5,i+n+1,c,x+y);		
	}
	printf("%lld\n",-flow::solve().second);
	return 0;
}
```