#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long ll;
const ll inf=1e16;
int n;
int S,T,cnt=1;
int head[2013];
struct Edge{
	int nxt,to;
	ll w,c;
}e[20003];
ll dis[2013],flow[2013];
int pre[2013],lst[2013];
bool v[2013];
void add(int u,int v,ll w,ll c)
{
	e[++cnt]=(Edge){head[u],v,w,c};
	head[u]=cnt;
	e[++cnt]=(Edge){head[v],u,0,-c};
	head[v]=cnt;
}
bool spfa()
{
	queue<int> q;
	for(int i=1;i<=T+4;i++)
		dis[i]=inf;
	dis[S]=0,flow[S]=inf;
	q.push(S);
	while(!q.empty())
	{
		int x=q.front();
		q.pop();
		v[x]=false;
		for(int i=head[x];i;i=e[i].nxt)
		{
			int to=e[i].to;
			if(e[i].w&&dis[to]>dis[x]+e[i].c)
			{
				dis[to]=dis[x]+e[i].c;
				flow[to]=min(flow[x],e[i].w);
				pre[to]=x,lst[to]=i;
				if(!v[to]) q.push(to),v[to]=true;
			}
		}
	}
	return dis[T]!=inf;
}
ll mcmf()
{
	ll ans=0;
	while(spfa())
	{
		ans+=dis[T]*flow[T];
		for(int i=T;i!=S;i=pre[i])
			e[lst[i]].w-=flow[T],e[lst[i]^1].w+=flow[T];
	}
	return ans;
}
int main()
{
	scanf("%d",&n);
	S=n+n+1,T=S+1;
	for(int i=1;i<=n;i++)
	{
		ll x,y,z;
		scanf("%lld%lld%lld",&x,&y,&z);
		add(S,i,z,0);
		add(i,T+1,inf,x+y);
		add(i,T+2,inf,-x-y);
		add(i,T+3,inf,x-y);
		add(i,T+4,inf,y-x);
	}
	for(int i=1;i<=n;i++)
	{
		ll x,y,z;
		scanf("%lld%lld%lld",&x,&y,&z);
		add(n+i,T,z,0);
		add(T+1,n+i,inf,-x-y);
		add(T+2,n+i,inf,x+y);
		add(T+3,n+i,inf,y-x);
		add(T+4,n+i,inf,x-y);
	}
	printf("%lld",-mcmf());
	return 0;
}
### Prompt

Your task is to create 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<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long ll;
const ll inf=1e16;
int n;
int S,T,cnt=1;
int head[2013];
struct Edge{
	int nxt,to;
	ll w,c;
}e[20003];
ll dis[2013],flow[2013];
int pre[2013],lst[2013];
bool v[2013];
void add(int u,int v,ll w,ll c)
{
	e[++cnt]=(Edge){head[u],v,w,c};
	head[u]=cnt;
	e[++cnt]=(Edge){head[v],u,0,-c};
	head[v]=cnt;
}
bool spfa()
{
	queue<int> q;
	for(int i=1;i<=T+4;i++)
		dis[i]=inf;
	dis[S]=0,flow[S]=inf;
	q.push(S);
	while(!q.empty())
	{
		int x=q.front();
		q.pop();
		v[x]=false;
		for(int i=head[x];i;i=e[i].nxt)
		{
			int to=e[i].to;
			if(e[i].w&&dis[to]>dis[x]+e[i].c)
			{
				dis[to]=dis[x]+e[i].c;
				flow[to]=min(flow[x],e[i].w);
				pre[to]=x,lst[to]=i;
				if(!v[to]) q.push(to),v[to]=true;
			}
		}
	}
	return dis[T]!=inf;
}
ll mcmf()
{
	ll ans=0;
	while(spfa())
	{
		ans+=dis[T]*flow[T];
		for(int i=T;i!=S;i=pre[i])
			e[lst[i]].w-=flow[T],e[lst[i]^1].w+=flow[T];
	}
	return ans;
}
int main()
{
	scanf("%d",&n);
	S=n+n+1,T=S+1;
	for(int i=1;i<=n;i++)
	{
		ll x,y,z;
		scanf("%lld%lld%lld",&x,&y,&z);
		add(S,i,z,0);
		add(i,T+1,inf,x+y);
		add(i,T+2,inf,-x-y);
		add(i,T+3,inf,x-y);
		add(i,T+4,inf,y-x);
	}
	for(int i=1;i<=n;i++)
	{
		ll x,y,z;
		scanf("%lld%lld%lld",&x,&y,&z);
		add(n+i,T,z,0);
		add(T+1,n+i,inf,-x-y);
		add(T+2,n+i,inf,x+y);
		add(T+3,n+i,inf,y-x);
		add(T+4,n+i,inf,x-y);
	}
	printf("%lld",-mcmf());
	return 0;
}
```