#include<cstdio>
#include<cstring>
using namespace std;
struct node
{
	int x,y,c,w,next,other;
}a[21000];int len,last[2100];
inline void ins(int x,int y,int c,int w)
{
	len++;a[len].x=x;a[len].y=y;a[len].c=c;a[len].w=w;
	a[len].next=last[x];last[x]=len;a[len].other=len+1;
	len++;a[len].x=y;a[len].y=x;a[len].c=0;a[len].w=-w;
	a[len].next=last[y];last[y]=len;a[len].other=len-1;
}
int n;
long long d[2100];
int pre[2100];
bool v[2100];
int list[2100];
void BellmanFord(int st)
{
	memset(d,-64,sizeof(d));d[st]=0;
	memset(v,false,sizeof(v));v[st]=true;
	int head=1,tail=2;list[1]=st;
	while(head!=tail)
	{
		int x=list[head];
		for(int k=last[x];k>0;k=a[k].next)
		{
			int y=a[k].y;
			if(a[k].c>0&&d[y]<d[x]+a[k].w)
			{
				d[y]=d[x]+a[k].w;
				pre[y]=k;
				if(v[y]==false)
				{
					v[y]=true;
					list[tail++]=y;
					if(tail==n*2+7)tail=1;
				}
			}
		}
		v[x]=false;
		head++;if(head==n*2+7)head=1;
	}
}
int main()
{
	scanf("%d",&n);
	int st=n*2+1,ed=n*2+6;
	len=0;memset(last,0,sizeof(last));
	for(int i=1;i<=n;i++)
	{
		int x,y,c;scanf("%d%d%d",&x,&y,&c);
		ins(st,i,c,0);
		ins(i,n*2+2,0x3f3f3f3f,x+y);
		ins(i,n*2+3,0x3f3f3f3f,-x+y);
		ins(i,n*2+4,0x3f3f3f3f,-x-y);
		ins(i,n*2+5,0x3f3f3f3f,x-y);
	}
	for(int i=1;i<=n;i++)
	{
		int x,y,c;scanf("%d%d%d",&x,&y,&c);
		ins(n+i,ed,c,0);
		ins(n*2+2,n+i,0x3f3f3f3f,-x-y);
		ins(n*2+3,n+i,0x3f3f3f3f,x-y);
		ins(n*2+4,n+i,0x3f3f3f3f,x+y);
		ins(n*2+5,n+i,0x3f3f3f3f,-x+y);
	}
	pre[st]=0;BellmanFord(st);
	long long ss=0;
	while(d[ed]!=0xc0c0c0c0c0c0c0c0ll)
	{
		int f=0x3f3f3f3f;
		for(int k=pre[ed];k>0;k=pre[a[k].x])if(a[k].c<f)f=a[k].c;
		for(int k=pre[ed];k>0;k=pre[a[k].x])a[k].c-=f,a[a[k].other].c+=f;
		ss+=d[ed]*f;
		BellmanFord(st);
	}
	printf("%lld\n",ss);
	return 0;
}
### Prompt

In Cpp, your task is to solve 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>
using namespace std;
struct node
{
	int x,y,c,w,next,other;
}a[21000];int len,last[2100];
inline void ins(int x,int y,int c,int w)
{
	len++;a[len].x=x;a[len].y=y;a[len].c=c;a[len].w=w;
	a[len].next=last[x];last[x]=len;a[len].other=len+1;
	len++;a[len].x=y;a[len].y=x;a[len].c=0;a[len].w=-w;
	a[len].next=last[y];last[y]=len;a[len].other=len-1;
}
int n;
long long d[2100];
int pre[2100];
bool v[2100];
int list[2100];
void BellmanFord(int st)
{
	memset(d,-64,sizeof(d));d[st]=0;
	memset(v,false,sizeof(v));v[st]=true;
	int head=1,tail=2;list[1]=st;
	while(head!=tail)
	{
		int x=list[head];
		for(int k=last[x];k>0;k=a[k].next)
		{
			int y=a[k].y;
			if(a[k].c>0&&d[y]<d[x]+a[k].w)
			{
				d[y]=d[x]+a[k].w;
				pre[y]=k;
				if(v[y]==false)
				{
					v[y]=true;
					list[tail++]=y;
					if(tail==n*2+7)tail=1;
				}
			}
		}
		v[x]=false;
		head++;if(head==n*2+7)head=1;
	}
}
int main()
{
	scanf("%d",&n);
	int st=n*2+1,ed=n*2+6;
	len=0;memset(last,0,sizeof(last));
	for(int i=1;i<=n;i++)
	{
		int x,y,c;scanf("%d%d%d",&x,&y,&c);
		ins(st,i,c,0);
		ins(i,n*2+2,0x3f3f3f3f,x+y);
		ins(i,n*2+3,0x3f3f3f3f,-x+y);
		ins(i,n*2+4,0x3f3f3f3f,-x-y);
		ins(i,n*2+5,0x3f3f3f3f,x-y);
	}
	for(int i=1;i<=n;i++)
	{
		int x,y,c;scanf("%d%d%d",&x,&y,&c);
		ins(n+i,ed,c,0);
		ins(n*2+2,n+i,0x3f3f3f3f,-x-y);
		ins(n*2+3,n+i,0x3f3f3f3f,x-y);
		ins(n*2+4,n+i,0x3f3f3f3f,x+y);
		ins(n*2+5,n+i,0x3f3f3f3f,-x+y);
	}
	pre[st]=0;BellmanFord(st);
	long long ss=0;
	while(d[ed]!=0xc0c0c0c0c0c0c0c0ll)
	{
		int f=0x3f3f3f3f;
		for(int k=pre[ed];k>0;k=pre[a[k].x])if(a[k].c<f)f=a[k].c;
		for(int k=pre[ed];k>0;k=pre[a[k].x])a[k].c-=f,a[a[k].other].c+=f;
		ss+=d[ed]*f;
		BellmanFord(st);
	}
	printf("%lld\n",ss);
	return 0;
}
```