#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<queue>
#include<cstring>
#define N 100010
#define int long long
using namespace std;
struct node{
	int nxt,to,w,f;
}road[N];
int head[N],cnt=1;
void add(int u,int v,int m,int w)
{
	road[++cnt]=(node){head[u],v,w,m};
	head[u]=cnt;
	road[++cnt]=(node){head[v],u,-w,0};
	head[v]=cnt;
}
int pre[N],bef[N],dis[N],flow[N];
bool in[N];
queue<int>q;
bool spfa(int s,int t)
{
	memset(dis,127,sizeof(dis));
	memset(flow,127,sizeof(flow));
	memset(in,0,sizeof(in));
	while(!q.empty())q.pop();
	q.push(s);
	dis[s]=0;
	in[s]=true;
	pre[t]=-1;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		in[u]=false;
		for(int i=head[u];i;i=road[i].nxt)
		{
			int v=road[i].to;
			if(road[i].f>0 && dis[v]>dis[u]+road[i].w)
			{
				dis[v]=dis[u]+road[i].w;
				pre[v]=u;
				bef[v]=i;
				flow[v]=min(flow[u],road[i].f);
				if(!in[v])
				{
					in[v]=true;
					q.push(v);
				}
			}
		}
	}
	return pre[t]!=-1;
}
void MCMF(int s,int t,int &maxf,int &minw)
{
	while(spfa(s,t))
	{
		int u=t;
		maxf+=flow[t];
		minw+=flow[t]*dis[t];
		while(u!=s)
		{
			road[bef[u]].f-=flow[t];
			road[bef[u]^1].f+=flow[t];
			u=pre[u];
		}
	}
}
signed main()
{
	int n;
	scanf("%d",&n);
	int s=2*n+5,t=s+1;
	int o1=2*n+1,o2=2*n+2,o3=2*n+3,o4=2*n+4;
	for(int i=1;i<=n;i++)
	{
		int x,y,w;
		scanf("%lld%lld%lld",&x,&y,&w);
		add(s,i,w,0);
		add(i,o1,w,x+y);
		add(i,o2,w,x-y);
		add(i,o3,w,-x+y);
		add(i,o4,w,-x-y);
	}
	
	for(int i=1;i<=n;i++)
	{
		int x,y,w;
		scanf("%lld%lld%lld",&x,&y,&w);
		add(i+n,t,w,0);
		add(o1,i+n,w,-x-y);
		add(o2,i+n,w,-x+y);
		add(o3,i+n,w,x-y);
		add(o4,i+n,w,x+y);
	}
	int ans=0,res=0;
	MCMF(s,t,res,ans);
	printf("%lld",-ans);
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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<cstdlib>
#include<iostream>
#include<queue>
#include<cstring>
#define N 100010
#define int long long
using namespace std;
struct node{
	int nxt,to,w,f;
}road[N];
int head[N],cnt=1;
void add(int u,int v,int m,int w)
{
	road[++cnt]=(node){head[u],v,w,m};
	head[u]=cnt;
	road[++cnt]=(node){head[v],u,-w,0};
	head[v]=cnt;
}
int pre[N],bef[N],dis[N],flow[N];
bool in[N];
queue<int>q;
bool spfa(int s,int t)
{
	memset(dis,127,sizeof(dis));
	memset(flow,127,sizeof(flow));
	memset(in,0,sizeof(in));
	while(!q.empty())q.pop();
	q.push(s);
	dis[s]=0;
	in[s]=true;
	pre[t]=-1;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		in[u]=false;
		for(int i=head[u];i;i=road[i].nxt)
		{
			int v=road[i].to;
			if(road[i].f>0 && dis[v]>dis[u]+road[i].w)
			{
				dis[v]=dis[u]+road[i].w;
				pre[v]=u;
				bef[v]=i;
				flow[v]=min(flow[u],road[i].f);
				if(!in[v])
				{
					in[v]=true;
					q.push(v);
				}
			}
		}
	}
	return pre[t]!=-1;
}
void MCMF(int s,int t,int &maxf,int &minw)
{
	while(spfa(s,t))
	{
		int u=t;
		maxf+=flow[t];
		minw+=flow[t]*dis[t];
		while(u!=s)
		{
			road[bef[u]].f-=flow[t];
			road[bef[u]^1].f+=flow[t];
			u=pre[u];
		}
	}
}
signed main()
{
	int n;
	scanf("%d",&n);
	int s=2*n+5,t=s+1;
	int o1=2*n+1,o2=2*n+2,o3=2*n+3,o4=2*n+4;
	for(int i=1;i<=n;i++)
	{
		int x,y,w;
		scanf("%lld%lld%lld",&x,&y,&w);
		add(s,i,w,0);
		add(i,o1,w,x+y);
		add(i,o2,w,x-y);
		add(i,o3,w,-x+y);
		add(i,o4,w,-x-y);
	}
	
	for(int i=1;i<=n;i++)
	{
		int x,y,w;
		scanf("%lld%lld%lld",&x,&y,&w);
		add(i+n,t,w,0);
		add(o1,i+n,w,-x-y);
		add(o2,i+n,w,-x+y);
		add(o3,i+n,w,x-y);
		add(o4,i+n,w,x+y);
	}
	int ans=0,res=0;
	MCMF(s,t,res,ans);
	printf("%lld",-ans);
	return 0;
}
```