#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 1010
#define ll long long
int last[N*2],cur[N*2],len=1;
ll dis[N*2];
int q[N*2],p[N*2];
ll ans=0;
struct
{
	int c,w,to,next;
}a[N*20];
int n,ns,m;
void add(int x,int y,int c,int w)
{
	a[++len].to=y;
	a[len].next=last[x];
	a[len].c=c,a[len].w=w;
	last[x]=len;
}
int id = 0;
int SPFA()
{
	for(int i=0;i<=ns;i++) dis[i]=-1e16;
	q[1]=0,p[0]=++id,dis[0]=0;
	int l=0,r=1;
	while(l!=r)
	{
		l=l%(ns+10)+1;
		int k=q[l];
		p[k]=0;
		for(int i=last[k];i;i=a[i].next) 
		{
			int x=a[i].to;
			if(dis[k]+a[i].w>dis[x]&&a[i].c)
			{
				dis[x]=dis[k]+a[i].w;
				if(p[x] < id)
				{
					p[x]=id;
					r=r%(ns+10)+1;
					q[r]=x;
				}
			}
		}
	}
	return dis[ns]>0;
}
int dfs(int k,int flow)
{
	if(k==ns) return flow;
	int have=0;
	p[k]=1;
	for(int i=cur[k];i;i=a[i].next)
	{
		int x=a[i].to;
		if(!p[x]&&dis[k]+a[i].w==dis[x]&&a[i].c)
		{
			cur[k] = i;
			int t=min(flow-have,a[i].c);
			int now=dfs(x,t);
			ans+=(ll)now*a[i].w;
			have+=now,a[i].c-=now,a[i^1].c+=now;
			if(have==flow) break;
		}
	}
	p[k]=0;
	return have;
}
int main()
{
	int i,x,y,c;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	{
		scanf("%d%d%d",&x,&y,&c);
		add(i,2*n+1,c,x+y);
		add(2*n+1,i,0,-x-y);
		add(i,2*n+2,c,x-y);
		add(2*n+2,i,0,y-x);
		add(i,2*n+3,c,y-x);
		add(2*n+3,i,0,x-y);
		add(i,2*n+4,c,-x-y);
		add(2*n+4,i,0,x+y);
		add(0,i,c,0);
		add(i,0,0,0);
	}
	for(i=1;i<=n;i++)
	{
		scanf("%d%d%d",&x,&y,&c);
		add(2*n+1,i+n,1e6,-x-y);
		add(i+n,2*n+1,0,x+y);
		add(2*n+2,i+n,1e6,y-x);
		add(i+n,2*n+2,0,x-y);
		add(2*n+3,i+n,1e6,x-y);
		add(i+n,2*n+3,0,y-x);
		add(2*n+4,i+n,1e6,x+y);
		add(i+n,2*n+4,0,-x-y);
		add(i+n,2*n+5,c,0);
		add(2*n+5,i+n,0,0);
	}
	ns=2*n+5;
	while(SPFA())
	{
		while(1) {
			for(i = 0; i <= ns; i++) cur[i] = last[i];
			int t = dfs(0,1e6);	
			if(!t) break;
		}
	}
	printf("%lld",ans);
	return 0;
}
### Prompt

Please create a solution in CPP 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<cstring>
#include<algorithm>
using namespace std;
#define N 1010
#define ll long long
int last[N*2],cur[N*2],len=1;
ll dis[N*2];
int q[N*2],p[N*2];
ll ans=0;
struct
{
	int c,w,to,next;
}a[N*20];
int n,ns,m;
void add(int x,int y,int c,int w)
{
	a[++len].to=y;
	a[len].next=last[x];
	a[len].c=c,a[len].w=w;
	last[x]=len;
}
int id = 0;
int SPFA()
{
	for(int i=0;i<=ns;i++) dis[i]=-1e16;
	q[1]=0,p[0]=++id,dis[0]=0;
	int l=0,r=1;
	while(l!=r)
	{
		l=l%(ns+10)+1;
		int k=q[l];
		p[k]=0;
		for(int i=last[k];i;i=a[i].next) 
		{
			int x=a[i].to;
			if(dis[k]+a[i].w>dis[x]&&a[i].c)
			{
				dis[x]=dis[k]+a[i].w;
				if(p[x] < id)
				{
					p[x]=id;
					r=r%(ns+10)+1;
					q[r]=x;
				}
			}
		}
	}
	return dis[ns]>0;
}
int dfs(int k,int flow)
{
	if(k==ns) return flow;
	int have=0;
	p[k]=1;
	for(int i=cur[k];i;i=a[i].next)
	{
		int x=a[i].to;
		if(!p[x]&&dis[k]+a[i].w==dis[x]&&a[i].c)
		{
			cur[k] = i;
			int t=min(flow-have,a[i].c);
			int now=dfs(x,t);
			ans+=(ll)now*a[i].w;
			have+=now,a[i].c-=now,a[i^1].c+=now;
			if(have==flow) break;
		}
	}
	p[k]=0;
	return have;
}
int main()
{
	int i,x,y,c;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	{
		scanf("%d%d%d",&x,&y,&c);
		add(i,2*n+1,c,x+y);
		add(2*n+1,i,0,-x-y);
		add(i,2*n+2,c,x-y);
		add(2*n+2,i,0,y-x);
		add(i,2*n+3,c,y-x);
		add(2*n+3,i,0,x-y);
		add(i,2*n+4,c,-x-y);
		add(2*n+4,i,0,x+y);
		add(0,i,c,0);
		add(i,0,0,0);
	}
	for(i=1;i<=n;i++)
	{
		scanf("%d%d%d",&x,&y,&c);
		add(2*n+1,i+n,1e6,-x-y);
		add(i+n,2*n+1,0,x+y);
		add(2*n+2,i+n,1e6,y-x);
		add(i+n,2*n+2,0,x-y);
		add(2*n+3,i+n,1e6,x-y);
		add(i+n,2*n+3,0,y-x);
		add(2*n+4,i+n,1e6,x+y);
		add(i+n,2*n+4,0,-x-y);
		add(i+n,2*n+5,c,0);
		add(2*n+5,i+n,0,0);
	}
	ns=2*n+5;
	while(SPFA())
	{
		while(1) {
			for(i = 0; i <= ns; i++) cur[i] = last[i];
			int t = dfs(0,1e6);	
			if(!t) break;
		}
	}
	printf("%lld",ans);
	return 0;
}
```