#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
struct Edge{
	int v,c,cost,nxt;
}e[50010];int tot=1;
int first[4010];
void build(int u,int v,int c,int cost)
{
	e[++tot]=(Edge){v,c,cost,first[u]};
	first[u]=tot;
	return ;
}
void adde(int u,int v,int c,int cost)
{
	build(u,v,c,cost);
	build(v,u,0,-cost);
	return ;
}
const long long INF=1e18;
int S,T;
long long dis[4010];
int pren[4010],pree[4010];
bool book[4010];
bool spfa()
{
	queue<int>q;
	for(int i=S;i<=T;i++)
		dis[i]=-INF;
	dis[S]=0;
	q.push(S);
	book[S]=true;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		book[u]=false;
		for(int i=first[u];i;i=e[i].nxt)
			if(e[i].c&&dis[u]+e[i].cost>dis[e[i].v])
			{
				dis[e[i].v]=dis[u]+e[i].cost;
				pren[e[i].v]=u;pree[e[i].v]=i;
				if(!book[e[i].v])
				{
					q.push(e[i].v);
					book[e[i].v]=true;
				}
			}
	}
	return dis[T]>-INF;
}
long long solve()
{
	long long ans=0;
	while(spfa())
	{
		int mn=1e9;
		for(int i=T;i!=S;i=pren[i])
		{
			int ee=pree[i];
			mn=min(mn,e[ee].c);
		}
		ans+=dis[T]*mn;
		for(int i=T;i!=S;i=pren[i])
		{
			int ee=pree[i];
			e[ee].c-=mn;
			e[ee^1].c+=mn;
		}
	}
	return ans;
}
int main()
{
	int n;scanf("%d",&n);
	S=0;T=n*2+5;
	for(int i=1;i<=n;i++)
	{
		int x,y,c;
		scanf("%d %d %d",&x,&y,&c);
		adde(S,i,c,0);
		adde(i,n*2+1,1e9,x+y);
		adde(i,n*2+2,1e9,x-y);
		adde(i,n*2+3,1e9,-x+y);
		adde(i,n*2+4,1e9,-x-y);
	}
	for(int i=1;i<=n;i++)
	{
		int x,y,c;
		scanf("%d %d %d",&x,&y,&c);
		adde(i+n,T,c,0);
		adde(n*2+1,i+n,1e9,-x-y);
		adde(n*2+2,i+n,1e9,-x+y);
		adde(n*2+3,i+n,1e9,x-y);
		adde(n*2+4,i+n,1e9,x+y);
	}
	printf("%lld\n",solve());
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
struct Edge{
	int v,c,cost,nxt;
}e[50010];int tot=1;
int first[4010];
void build(int u,int v,int c,int cost)
{
	e[++tot]=(Edge){v,c,cost,first[u]};
	first[u]=tot;
	return ;
}
void adde(int u,int v,int c,int cost)
{
	build(u,v,c,cost);
	build(v,u,0,-cost);
	return ;
}
const long long INF=1e18;
int S,T;
long long dis[4010];
int pren[4010],pree[4010];
bool book[4010];
bool spfa()
{
	queue<int>q;
	for(int i=S;i<=T;i++)
		dis[i]=-INF;
	dis[S]=0;
	q.push(S);
	book[S]=true;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		book[u]=false;
		for(int i=first[u];i;i=e[i].nxt)
			if(e[i].c&&dis[u]+e[i].cost>dis[e[i].v])
			{
				dis[e[i].v]=dis[u]+e[i].cost;
				pren[e[i].v]=u;pree[e[i].v]=i;
				if(!book[e[i].v])
				{
					q.push(e[i].v);
					book[e[i].v]=true;
				}
			}
	}
	return dis[T]>-INF;
}
long long solve()
{
	long long ans=0;
	while(spfa())
	{
		int mn=1e9;
		for(int i=T;i!=S;i=pren[i])
		{
			int ee=pree[i];
			mn=min(mn,e[ee].c);
		}
		ans+=dis[T]*mn;
		for(int i=T;i!=S;i=pren[i])
		{
			int ee=pree[i];
			e[ee].c-=mn;
			e[ee^1].c+=mn;
		}
	}
	return ans;
}
int main()
{
	int n;scanf("%d",&n);
	S=0;T=n*2+5;
	for(int i=1;i<=n;i++)
	{
		int x,y,c;
		scanf("%d %d %d",&x,&y,&c);
		adde(S,i,c,0);
		adde(i,n*2+1,1e9,x+y);
		adde(i,n*2+2,1e9,x-y);
		adde(i,n*2+3,1e9,-x+y);
		adde(i,n*2+4,1e9,-x-y);
	}
	for(int i=1;i<=n;i++)
	{
		int x,y,c;
		scanf("%d %d %d",&x,&y,&c);
		adde(i+n,T,c,0);
		adde(n*2+1,i+n,1e9,-x-y);
		adde(n*2+2,i+n,1e9,-x+y);
		adde(n*2+3,i+n,1e9,x-y);
		adde(n*2+4,i+n,1e9,x+y);
	}
	printf("%lld\n",solve());
	return 0;
}
```