#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#define ll long long
using namespace std;
const int INF=2147483647;
struct Edge
{
	int to;
	int nxt;
	int flow;
	ll cost;
}e[50005];
struct Point
{
	ll x,y;
	int c;
}p1[1005],p2[1005];
int n,m,s,t,edgenum=1,head[5005],a[5005],pre[5005];
ll dis[5005];
bool flag[5005];
queue<int>q;
void add(int u,int v,int f,ll c)
{
	e[++edgenum].cost=c;
	e[edgenum].flow=f;
	e[edgenum].to=v;
	e[edgenum].nxt=head[u];
	head[u]=edgenum;
}
bool SPFA()
{
	for(int i=1;i<=t;i++)dis[i]=1e18,a[i]=0;
	dis[s]=0,a[s]=INF;
	flag[s]=1;
	q.push(s);
	while(!q.empty())
	{
		int node=q.front();
		q.pop();
		flag[node]=0;
		for(int hd=head[node];hd;hd=e[hd].nxt)
		{
			int to=e[hd].to;
			if(e[hd].flow==0)continue;
			if(dis[to]>dis[node]+e[hd].cost)
			{
				dis[to]=dis[node]+e[hd].cost;
				a[to]=min(a[node],e[hd].flow);
				pre[to]=hd;
				if(!flag[to])
				{
					q.push(to);
					flag[to]=1;
				}
			}
		}
	}
	return dis[t]<1e17;
}
ll EK()
{
	ll ans=0;
	while(SPFA())
	{
		ans+=dis[t]*a[t];
		int x=t;
		while(x)
		{
			int i=pre[x];
			e[i].flow-=a[t];
			e[i^1].flow+=a[t];
			x=e[i^1].to;
		}
	}
	return ans;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		ll x,y;
		scanf("%lld%lld%d",&x,&y,&p1[i].c);
		p1[i].x=x+y,p1[i].y=x-y;
	}
	for(int i=1;i<=n;i++)
	{
		ll x,y;
		scanf("%lld%lld%d",&x,&y,&p2[i].c);
		p2[i].x=x+y,p2[i].y=x-y;
	}
	s=2*n+5,t=2*n+6;
	for(int i=1;i<=n;i++)
	{
		add(2*n+1,i+n,INF,-p2[i].x);
		add(i+n,2*n+1,0,p2[i].x);
		add(2*n+2,i+n,INF,p2[i].x);
		add(i+n,2*n+2,0,-p2[i].x);
		add(2*n+3,i+n,INF,-p2[i].y);
		add(i+n,2*n+3,0,p2[i].y);
		add(2*n+4,i+n,INF,p2[i].y);
		add(i+n,2*n+4,0,-p2[i].y);
		add(s,i,p1[i].c,0);
		add(i,s,0,0);
		add(i+n,t,p2[i].c,0);
		add(t,i+n,0,0);
		add(i,2*n+1,INF,p1[i].x);
		add(2*n+1,i,0,-p1[i].x);
		add(i,2*n+2,INF,-p1[i].x);
		add(2*n+2,i,0,p1[i].x);
		add(i,2*n+3,INF,p1[i].y);
		add(2*n+3,i,0,-p1[i].y);
		add(i,2*n+4,INF,-p1[i].y);
		add(2*n+4,i,0,p1[i].y);
	}
	printf("%lld\n",-EK());
	return 0;
}
### Prompt

Your challenge is to write 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<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#define ll long long
using namespace std;
const int INF=2147483647;
struct Edge
{
	int to;
	int nxt;
	int flow;
	ll cost;
}e[50005];
struct Point
{
	ll x,y;
	int c;
}p1[1005],p2[1005];
int n,m,s,t,edgenum=1,head[5005],a[5005],pre[5005];
ll dis[5005];
bool flag[5005];
queue<int>q;
void add(int u,int v,int f,ll c)
{
	e[++edgenum].cost=c;
	e[edgenum].flow=f;
	e[edgenum].to=v;
	e[edgenum].nxt=head[u];
	head[u]=edgenum;
}
bool SPFA()
{
	for(int i=1;i<=t;i++)dis[i]=1e18,a[i]=0;
	dis[s]=0,a[s]=INF;
	flag[s]=1;
	q.push(s);
	while(!q.empty())
	{
		int node=q.front();
		q.pop();
		flag[node]=0;
		for(int hd=head[node];hd;hd=e[hd].nxt)
		{
			int to=e[hd].to;
			if(e[hd].flow==0)continue;
			if(dis[to]>dis[node]+e[hd].cost)
			{
				dis[to]=dis[node]+e[hd].cost;
				a[to]=min(a[node],e[hd].flow);
				pre[to]=hd;
				if(!flag[to])
				{
					q.push(to);
					flag[to]=1;
				}
			}
		}
	}
	return dis[t]<1e17;
}
ll EK()
{
	ll ans=0;
	while(SPFA())
	{
		ans+=dis[t]*a[t];
		int x=t;
		while(x)
		{
			int i=pre[x];
			e[i].flow-=a[t];
			e[i^1].flow+=a[t];
			x=e[i^1].to;
		}
	}
	return ans;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		ll x,y;
		scanf("%lld%lld%d",&x,&y,&p1[i].c);
		p1[i].x=x+y,p1[i].y=x-y;
	}
	for(int i=1;i<=n;i++)
	{
		ll x,y;
		scanf("%lld%lld%d",&x,&y,&p2[i].c);
		p2[i].x=x+y,p2[i].y=x-y;
	}
	s=2*n+5,t=2*n+6;
	for(int i=1;i<=n;i++)
	{
		add(2*n+1,i+n,INF,-p2[i].x);
		add(i+n,2*n+1,0,p2[i].x);
		add(2*n+2,i+n,INF,p2[i].x);
		add(i+n,2*n+2,0,-p2[i].x);
		add(2*n+3,i+n,INF,-p2[i].y);
		add(i+n,2*n+3,0,p2[i].y);
		add(2*n+4,i+n,INF,p2[i].y);
		add(i+n,2*n+4,0,-p2[i].y);
		add(s,i,p1[i].c,0);
		add(i,s,0,0);
		add(i+n,t,p2[i].c,0);
		add(t,i+n,0,0);
		add(i,2*n+1,INF,p1[i].x);
		add(2*n+1,i,0,-p1[i].x);
		add(i,2*n+2,INF,-p1[i].x);
		add(2*n+2,i,0,p1[i].x);
		add(i,2*n+3,INF,p1[i].y);
		add(2*n+3,i,0,-p1[i].y);
		add(i,2*n+4,INF,-p1[i].y);
		add(2*n+4,i,0,p1[i].y);
	}
	printf("%lld\n",-EK());
	return 0;
}
```