#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 2050
struct edge{int t,next,v,c;}ed[N*30];
int n,a,b,c,head[N],ct,cnt,vis2[N],vis3[N],vis[N];
long long dis[N],as;
void adde(int f,int t,int w,int c){ed[++cnt]=(edge){t,head[f],w,c};head[f]=cnt;ed[++cnt]=(edge){f,head[t],0,-c};head[t]=cnt;}
void bfs2(int t)
{
	memset(vis2,0,sizeof(vis2));
	queue<int> tp;
	tp.push(t);
	vis2[t]=1;
	while(!tp.empty())
	{
		int x=tp.front();tp.pop();
		for(int i=head[x];i;i=ed[i].next)
		if(ed[i^1].v)
		if(!vis2[ed[i].t])vis2[ed[i].t]=1,tp.push(ed[i].t);
	}
}
bool spfa_is_dead(int s,int t)
{
	bfs2(t);
	memset(vis,0,sizeof(vis));
	memset(vis3,0,sizeof(vis3));
	memset(dis,0x3f,sizeof(dis));
	queue<int> tp;
	tp.push(s);
	dis[s]=0;vis[s]=1;
	while(!tp.empty())
	{
		int x=tp.front();tp.pop();
		vis[x]=0;
		for(int i=head[x];i;i=ed[i].next)
		if(ed[i].v&&vis2[ed[i].t])
		if(dis[ed[i].t]>dis[x]+ed[i].c)
		{
			dis[ed[i].t]=dis[x]+ed[i].c;
			if(!vis[ed[i].t])vis[ed[i].t]=1,tp.push(ed[i].t);
		}
	}
	return dis[t]<=1e15;
}
int dfs(int u,int t,int f)
{
	if(u==t||!f)return f;
	vis3[u]=1;
	int as1=0,tp;
	for(int i=head[u];i;i=ed[i].next)
	if(!vis3[ed[i].t]&&vis2[ed[i].t]&&ed[i].v&&dis[ed[i].t]==dis[u]+ed[i].c&&(tp=dfs(ed[i].t,t,min(ed[i].v,f))))
	{
		as1+=tp,f-=tp;
		ed[i].v-=tp,ed[i^1].v+=tp;
		as+=1ll*ed[i].c*tp;
		if(!f)return as1;
	}
	return as1;
}
long long dinic(int s,int t)
{
	while(spfa_is_dead(s,t))
	dfs(s,t,1e8);
	return as;
}
int main()
{
	scanf("%d",&n);
	ct=6;cnt=1;
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d%d",&a,&b,&c);
		++ct;
		adde(1,ct,c,0);
		adde(ct,3,c,a+b);
		adde(ct,4,c,a-b);
		adde(ct,5,c,-a+b);
		adde(ct,6,c,-a-b);
	}
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d%d",&a,&b,&c);
		++ct;
		adde(ct,2,c,0);
		adde(3,ct,c,-a-b);
		adde(4,ct,c,-a+b);
		adde(5,ct,c,a-b);
		adde(6,ct,c,a+b);
	}
	printf("%lld\n",-dinic(1,2));
}
### Prompt

Generate 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
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 2050
struct edge{int t,next,v,c;}ed[N*30];
int n,a,b,c,head[N],ct,cnt,vis2[N],vis3[N],vis[N];
long long dis[N],as;
void adde(int f,int t,int w,int c){ed[++cnt]=(edge){t,head[f],w,c};head[f]=cnt;ed[++cnt]=(edge){f,head[t],0,-c};head[t]=cnt;}
void bfs2(int t)
{
	memset(vis2,0,sizeof(vis2));
	queue<int> tp;
	tp.push(t);
	vis2[t]=1;
	while(!tp.empty())
	{
		int x=tp.front();tp.pop();
		for(int i=head[x];i;i=ed[i].next)
		if(ed[i^1].v)
		if(!vis2[ed[i].t])vis2[ed[i].t]=1,tp.push(ed[i].t);
	}
}
bool spfa_is_dead(int s,int t)
{
	bfs2(t);
	memset(vis,0,sizeof(vis));
	memset(vis3,0,sizeof(vis3));
	memset(dis,0x3f,sizeof(dis));
	queue<int> tp;
	tp.push(s);
	dis[s]=0;vis[s]=1;
	while(!tp.empty())
	{
		int x=tp.front();tp.pop();
		vis[x]=0;
		for(int i=head[x];i;i=ed[i].next)
		if(ed[i].v&&vis2[ed[i].t])
		if(dis[ed[i].t]>dis[x]+ed[i].c)
		{
			dis[ed[i].t]=dis[x]+ed[i].c;
			if(!vis[ed[i].t])vis[ed[i].t]=1,tp.push(ed[i].t);
		}
	}
	return dis[t]<=1e15;
}
int dfs(int u,int t,int f)
{
	if(u==t||!f)return f;
	vis3[u]=1;
	int as1=0,tp;
	for(int i=head[u];i;i=ed[i].next)
	if(!vis3[ed[i].t]&&vis2[ed[i].t]&&ed[i].v&&dis[ed[i].t]==dis[u]+ed[i].c&&(tp=dfs(ed[i].t,t,min(ed[i].v,f))))
	{
		as1+=tp,f-=tp;
		ed[i].v-=tp,ed[i^1].v+=tp;
		as+=1ll*ed[i].c*tp;
		if(!f)return as1;
	}
	return as1;
}
long long dinic(int s,int t)
{
	while(spfa_is_dead(s,t))
	dfs(s,t,1e8);
	return as;
}
int main()
{
	scanf("%d",&n);
	ct=6;cnt=1;
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d%d",&a,&b,&c);
		++ct;
		adde(1,ct,c,0);
		adde(ct,3,c,a+b);
		adde(ct,4,c,a-b);
		adde(ct,5,c,-a+b);
		adde(ct,6,c,-a-b);
	}
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d%d",&a,&b,&c);
		++ct;
		adde(ct,2,c,0);
		adde(3,ct,c,-a-b);
		adde(4,ct,c,-a+b);
		adde(5,ct,c,a-b);
		adde(6,ct,c,a+b);
	}
	printf("%lld\n",-dinic(1,2));
}
```