#include <bits/stdc++.h>
#define fo(i,a,b) for(int i=a;i<=b;++i)
#define fod(i,a,b) for(int i=a;i>=b;--i)
const int N=2115;
const int INF=1e7;
typedef long long LL;
using namespace std;
vector <int> ap[N];
int n,n1,st,ed,f[N][N];
LL ans,pr[N][N];
void link(int x,int y,int w,LL c)
{
	ap[x].push_back(y);
	f[x][y]=w,pr[x][y]=c;
	ap[y].push_back(x);
	f[y][x]=0,pr[y][x]=-c;
}
typedef vector<int>::iterator IT;
namespace Flow
{
	LL dis[N];
	bool bz[N];
	IT cur[N];
	int d[200*N];
	bool spfa()
	{
		memset(dis,107,sizeof(dis));
		memset(bz,0,sizeof(bz));
		dis[st]=0,bz[st]=1,d[1]=st;
		fo(i,1,n1) cur[i]=ap[i].begin();
		int l=0,r=1;
		while(l<r)
		{
			int k=d[++l];
			for(IT i=ap[k].begin();i!=ap[k].end();i++)	
			{
				int p=*i;
				if(f[k][p]&&dis[k]+pr[k][p]<dis[p]) 
				{
					dis[p]=dis[k]+pr[k][p];
					if(!bz[p]) bz[p]=1,d[++r]=p;
				}
			}
			bz[k]=0;
		}
		return (dis[ed]<=1e17);
	}
	int flow(int k,int s)
	{
		if(k==ed) return s;
		int sl=0,v;
		bz[k]=1;
		for(;cur[k]!=ap[k].end();cur[k]++)
		{
			int p=*cur[k];
			if(!bz[p]&&f[k][p]&&dis[p]==dis[k]+pr[k][p]) 
			{
				if(v=flow(p,min(s,f[k][p]))) 
				{
					sl+=v,s-=v;
					f[k][p]-=v,f[p][k]+=v;
					ans+=(LL)v*pr[k][p];
					if(!s) break;
				}
			}
		}
		bz[k]=0;
		return sl;
	}
}
using Flow::flow;
using Flow::spfa;
int main()
{	
	cin>>n;
	n1=2*n+6,st=2*n+5,ed=n1;
	fo(i,1,n) 
	{
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		link(st,i,z,0);
		link(i,2*n+1,z,x+y);
		link(i,2*n+2,z,x-y);
		link(i,2*n+3,z,-x+y);
		link(i,2*n+4,z,-x-y);
	}  
	fo(i,1,n) 
	{
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		link(i+n,ed,z,0);
		link(2*n+1,i+n,z,-x-y);
		link(2*n+2,i+n,z,-x+y);
		link(2*n+3,i+n,z,x-y);
		link(2*n+4,i+n,z,x+y);
	}
	ans=0;
	while(spfa()) 
		flow(st,INF);
	printf("%lld\n",-ans);
}
### Prompt

Create a solution in Cpp for 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 <bits/stdc++.h>
#define fo(i,a,b) for(int i=a;i<=b;++i)
#define fod(i,a,b) for(int i=a;i>=b;--i)
const int N=2115;
const int INF=1e7;
typedef long long LL;
using namespace std;
vector <int> ap[N];
int n,n1,st,ed,f[N][N];
LL ans,pr[N][N];
void link(int x,int y,int w,LL c)
{
	ap[x].push_back(y);
	f[x][y]=w,pr[x][y]=c;
	ap[y].push_back(x);
	f[y][x]=0,pr[y][x]=-c;
}
typedef vector<int>::iterator IT;
namespace Flow
{
	LL dis[N];
	bool bz[N];
	IT cur[N];
	int d[200*N];
	bool spfa()
	{
		memset(dis,107,sizeof(dis));
		memset(bz,0,sizeof(bz));
		dis[st]=0,bz[st]=1,d[1]=st;
		fo(i,1,n1) cur[i]=ap[i].begin();
		int l=0,r=1;
		while(l<r)
		{
			int k=d[++l];
			for(IT i=ap[k].begin();i!=ap[k].end();i++)	
			{
				int p=*i;
				if(f[k][p]&&dis[k]+pr[k][p]<dis[p]) 
				{
					dis[p]=dis[k]+pr[k][p];
					if(!bz[p]) bz[p]=1,d[++r]=p;
				}
			}
			bz[k]=0;
		}
		return (dis[ed]<=1e17);
	}
	int flow(int k,int s)
	{
		if(k==ed) return s;
		int sl=0,v;
		bz[k]=1;
		for(;cur[k]!=ap[k].end();cur[k]++)
		{
			int p=*cur[k];
			if(!bz[p]&&f[k][p]&&dis[p]==dis[k]+pr[k][p]) 
			{
				if(v=flow(p,min(s,f[k][p]))) 
				{
					sl+=v,s-=v;
					f[k][p]-=v,f[p][k]+=v;
					ans+=(LL)v*pr[k][p];
					if(!s) break;
				}
			}
		}
		bz[k]=0;
		return sl;
	}
}
using Flow::flow;
using Flow::spfa;
int main()
{	
	cin>>n;
	n1=2*n+6,st=2*n+5,ed=n1;
	fo(i,1,n) 
	{
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		link(st,i,z,0);
		link(i,2*n+1,z,x+y);
		link(i,2*n+2,z,x-y);
		link(i,2*n+3,z,-x+y);
		link(i,2*n+4,z,-x-y);
	}  
	fo(i,1,n) 
	{
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		link(i+n,ed,z,0);
		link(2*n+1,i+n,z,-x-y);
		link(2*n+2,i+n,z,-x+y);
		link(2*n+3,i+n,z,x-y);
		link(2*n+4,i+n,z,x+y);
	}
	ans=0;
	while(spfa()) 
		flow(st,INF);
	printf("%lld\n",-ans);
}
```