#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
#define inf (1ll<<61)
#define MAX 2015
using namespace std;
//char nc()
//{
//	static char buf[100000],*p1=buf,*p2=buf;
//	return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
//}
char nc(){return getchar();}
int read()
{
	int x=0,y=1;
	char c=nc();
	while(!isdigit(c))
	{
		if(c=='-')y=-1;
		c=nc();
	}
	while(isdigit(c))
	{
		x=(x<<1)+(x<<3)+(c^48);
		c=nc();
	}
	return x*y;
}
int n,head[MAX],last[MAX],cnt=1,S,T,NUM;
struct edge{
	int next,to,flow;
	long long cost;
}a[MAX*24];
void add(int x,int y,int z,long long c)
{
	a[++cnt].next=head[x];
	a[cnt].to=y;
	a[cnt].flow=z;
	a[cnt].cost=c;
	head[x]=cnt;
	a[++cnt].next=head[y];
	a[cnt].to=x;
	a[cnt].flow=0;
	a[cnt].cost=-c;
	head[y]=cnt;
}
long long d[MAX],ANS=0;
int inqueue[MAX],vst[MAX];
bool spfa()
{
	queue<int> q;
	for(int i=1;i<=NUM;i++)
		d[i]=-inf;
	d[T]=0;inqueue[T]=1;
	q.push(T);
	while(!q.empty())
	{
		int x=q.front();
		q.pop();
		inqueue[x]=0;
		for(int i=head[x];i;i=a[i].next)
		{
			int t=a[i].to;
			if(a[i^1].flow&&d[t]<d[x]+a[i^1].cost)
			{
				d[t]=d[x]+a[i^1].cost;
				if(!inqueue[t])
				{
					inqueue[t]=1;
					q.push(t);
				}
			}
		}
	}
	return d[S]>-inf;
}
int dfs(int x,int maxf)
{
	vst[x]=1;
	if(x==T) return maxf;
	int used=0,dlt;
	for(int i=last[x];i;i=a[i].next)
	{
		int t=a[i].to;
		if(!vst[t]&&a[i].flow&&d[x]==d[t]+a[i].cost)
		{
			dlt=dfs(t,min(maxf-used,a[i].flow));
			ANS+=a[i].cost*dlt;
			a[i].flow-=dlt;
			a[i^1].flow+=dlt;
			used+=dlt;
			if(used==maxf)
			{
				vst[x]=0;
				break;
			}
		}
		last[x]=i;
	}
	return used;
}
int zkw()
{
	int ans=0;
	while(spfa())
	{
		memset(vst,0,sizeof(vst));
		memcpy(last,head,sizeof(head));
		vst[T]=1;
		while(vst[T])
		{
			vst[T]=0;
			ans+=dfs(S,INF);
		}
	}
	return ans;
}
int main()
{
	n=read();
	NUM=n*2;S=++NUM;T=++NUM;
	int x,y,z;
	for(int i=1;i<=n;i++)
	{
		x=read();y=read();z=read();
		add(S,i,z,0);
		add(i,NUM+1,INF,x+y);
		add(i,NUM+2,INF,-x+y);
		add(i,NUM+3,INF,x-y);
		add(i,NUM+4,INF,-x-y);
	}
	for(int i=1;i<=n;i++)
	{
		x=read();y=read();z=read();
		add(i+n,T,z,0);
		add(NUM+1,i+n,INF,-x-y);
		add(NUM+2,i+n,INF,x-y);
		add(NUM+3,i+n,INF,-x+y);
		add(NUM+4,i+n,INF,x+y);
	}
	NUM+=4;
	zkw();
	printf("%lld",ANS);
	return 0;
}

### 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
#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
#define inf (1ll<<61)
#define MAX 2015
using namespace std;
//char nc()
//{
//	static char buf[100000],*p1=buf,*p2=buf;
//	return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
//}
char nc(){return getchar();}
int read()
{
	int x=0,y=1;
	char c=nc();
	while(!isdigit(c))
	{
		if(c=='-')y=-1;
		c=nc();
	}
	while(isdigit(c))
	{
		x=(x<<1)+(x<<3)+(c^48);
		c=nc();
	}
	return x*y;
}
int n,head[MAX],last[MAX],cnt=1,S,T,NUM;
struct edge{
	int next,to,flow;
	long long cost;
}a[MAX*24];
void add(int x,int y,int z,long long c)
{
	a[++cnt].next=head[x];
	a[cnt].to=y;
	a[cnt].flow=z;
	a[cnt].cost=c;
	head[x]=cnt;
	a[++cnt].next=head[y];
	a[cnt].to=x;
	a[cnt].flow=0;
	a[cnt].cost=-c;
	head[y]=cnt;
}
long long d[MAX],ANS=0;
int inqueue[MAX],vst[MAX];
bool spfa()
{
	queue<int> q;
	for(int i=1;i<=NUM;i++)
		d[i]=-inf;
	d[T]=0;inqueue[T]=1;
	q.push(T);
	while(!q.empty())
	{
		int x=q.front();
		q.pop();
		inqueue[x]=0;
		for(int i=head[x];i;i=a[i].next)
		{
			int t=a[i].to;
			if(a[i^1].flow&&d[t]<d[x]+a[i^1].cost)
			{
				d[t]=d[x]+a[i^1].cost;
				if(!inqueue[t])
				{
					inqueue[t]=1;
					q.push(t);
				}
			}
		}
	}
	return d[S]>-inf;
}
int dfs(int x,int maxf)
{
	vst[x]=1;
	if(x==T) return maxf;
	int used=0,dlt;
	for(int i=last[x];i;i=a[i].next)
	{
		int t=a[i].to;
		if(!vst[t]&&a[i].flow&&d[x]==d[t]+a[i].cost)
		{
			dlt=dfs(t,min(maxf-used,a[i].flow));
			ANS+=a[i].cost*dlt;
			a[i].flow-=dlt;
			a[i^1].flow+=dlt;
			used+=dlt;
			if(used==maxf)
			{
				vst[x]=0;
				break;
			}
		}
		last[x]=i;
	}
	return used;
}
int zkw()
{
	int ans=0;
	while(spfa())
	{
		memset(vst,0,sizeof(vst));
		memcpy(last,head,sizeof(head));
		vst[T]=1;
		while(vst[T])
		{
			vst[T]=0;
			ans+=dfs(S,INF);
		}
	}
	return ans;
}
int main()
{
	n=read();
	NUM=n*2;S=++NUM;T=++NUM;
	int x,y,z;
	for(int i=1;i<=n;i++)
	{
		x=read();y=read();z=read();
		add(S,i,z,0);
		add(i,NUM+1,INF,x+y);
		add(i,NUM+2,INF,-x+y);
		add(i,NUM+3,INF,x-y);
		add(i,NUM+4,INF,-x-y);
	}
	for(int i=1;i<=n;i++)
	{
		x=read();y=read();z=read();
		add(i+n,T,z,0);
		add(NUM+1,i+n,INF,-x-y);
		add(NUM+2,i+n,INF,x-y);
		add(NUM+3,i+n,INF,-x+y);
		add(NUM+4,i+n,INF,x+y);
	}
	NUM+=4;
	zkw();
	printf("%lld",ANS);
	return 0;
}

```