#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=2115;
const ll inf=1e17;
int n,tt,cc,S,T,hd[N],iter[N],vis[N];ll ans,dis[N];queue<int>q;
struct E{int v,nxt,w;ll c;}e[N*10];
inline void add(int u,int v,int w,ll c){e[++cc]=(E){v,hd[u],w,c};hd[u]=cc;e[++cc]=(E){u,hd[v],0,-c};hd[v]=cc;}
inline bool spfa()
{
	for(int i=1;i<=tt;i++)iter[i]=hd[i],dis[i]=inf;
	dis[S]=0;vis[S]=1;q.push(S);
	while(!q.empty())
	{
		int x=q.front();q.pop();vis[x]=0;
		for(int i=hd[x];i;i=e[i].nxt)
		{
			int y=e[i].v;
			if(e[i].w&&dis[y]>dis[x]+e[i].c)
			{
				dis[y]=dis[x]+e[i].c;
				if(!vis[y]){vis[y]=1;q.push(y);}
			}
		}
	}
	return dis[T]<inf;
}
int dfs(int x,int f)
{
	if(x==T||!f)return f;
	int sf=0;vis[x]=1;
	for(int &i=iter[x];i;i=e[i].nxt)
	{
		int y=e[i].v;if(vis[y]||dis[y]!=dis[x]+e[i].c)continue;
		int w=dfs(y,min(f,e[i].w));
		if(w){e[i].w-=w;e[i^1].w+=w;sf+=w;f-=w;ans+=1ll*w*e[i].c;}
		if(!f)break;
	}
	vis[x]=0;return sf;
}
int main()
{
	scanf("%d",&n);tt=2*n+6;S=tt-1;T=tt;int t1=2*n+1,t2=t1+1,t3=t2+1,t4=t3+1;cc=1;
	for(int i=1,x,y,w;i<=n;i++){scanf("%d%d%d",&x,&y,&w);add(S,i,w,0);add(i,t1,w,x+y);add(i,t2,w,x-y);add(i,t3,w,-x+y);add(i,t4,w,-x-y);}
	for(int i=1,x,y,w;i<=n;i++){int t=n+i;scanf("%d%d%d",&x,&y,&w);add(t,T,w,0);add(t1,t,w,-x-y);add(t2,t,w,-x+y);add(t3,t,w,x-y);add(t4,t,w,x+y);}
	ans=0;while(spfa())dfs(S,1e7);printf("%lld\n",-ans);return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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>
using namespace std;
#define ll long long
const int N=2115;
const ll inf=1e17;
int n,tt,cc,S,T,hd[N],iter[N],vis[N];ll ans,dis[N];queue<int>q;
struct E{int v,nxt,w;ll c;}e[N*10];
inline void add(int u,int v,int w,ll c){e[++cc]=(E){v,hd[u],w,c};hd[u]=cc;e[++cc]=(E){u,hd[v],0,-c};hd[v]=cc;}
inline bool spfa()
{
	for(int i=1;i<=tt;i++)iter[i]=hd[i],dis[i]=inf;
	dis[S]=0;vis[S]=1;q.push(S);
	while(!q.empty())
	{
		int x=q.front();q.pop();vis[x]=0;
		for(int i=hd[x];i;i=e[i].nxt)
		{
			int y=e[i].v;
			if(e[i].w&&dis[y]>dis[x]+e[i].c)
			{
				dis[y]=dis[x]+e[i].c;
				if(!vis[y]){vis[y]=1;q.push(y);}
			}
		}
	}
	return dis[T]<inf;
}
int dfs(int x,int f)
{
	if(x==T||!f)return f;
	int sf=0;vis[x]=1;
	for(int &i=iter[x];i;i=e[i].nxt)
	{
		int y=e[i].v;if(vis[y]||dis[y]!=dis[x]+e[i].c)continue;
		int w=dfs(y,min(f,e[i].w));
		if(w){e[i].w-=w;e[i^1].w+=w;sf+=w;f-=w;ans+=1ll*w*e[i].c;}
		if(!f)break;
	}
	vis[x]=0;return sf;
}
int main()
{
	scanf("%d",&n);tt=2*n+6;S=tt-1;T=tt;int t1=2*n+1,t2=t1+1,t3=t2+1,t4=t3+1;cc=1;
	for(int i=1,x,y,w;i<=n;i++){scanf("%d%d%d",&x,&y,&w);add(S,i,w,0);add(i,t1,w,x+y);add(i,t2,w,x-y);add(i,t3,w,-x+y);add(i,t4,w,-x-y);}
	for(int i=1,x,y,w;i<=n;i++){int t=n+i;scanf("%d%d%d",&x,&y,&w);add(t,T,w,0);add(t1,t,w,-x-y);add(t2,t,w,-x+y);add(t3,t,w,x-y);add(t4,t,w,x+y);}
	ans=0;while(spfa())dfs(S,1e7);printf("%lld\n",-ans);return 0;
}
```