#include<bits/stdc++.h>
#define ll long long
#define ull unsigned ll
#define uint unsigned
#define pii pair<int,int>
#define pll pair<ll,ll>
#define PB push_back
#define fi first
#define se second
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define CLR(a,v) memset(a,v,sizeof(a));
#define CPY(a,b) memcpy(a,b,sizeof(a));
using namespace std;
const int N=2010;
struct edge{
	int from,to,next,f,v;
}e[N*10];
int head[N],tot=1;
int n,nd,S,T,p0,p1,p2,p3;
int vis[N],q[N*10],from[N];
ll dis[N],ans;
void add(int x,int y,int f,int v){
	e[++tot]=(edge){x,y,head[x],f,v};
	head[x]=tot;
	e[++tot]=(edge){y,x,head[y],0,-v};
	head[y]=tot;
}
bool bfs(){
	For(i,1,nd){
		vis[i]=0;
		dis[i]=-(1ll<<60);
	}
	int h=0,t=1;
	q[1]=S; dis[S]=0;
	while (h!=t){
		int x=q[++h]; vis[x]=0;
		for (int i=head[x];i;i=e[i].next)
			if (dis[e[i].to]<dis[x]+e[i].v&&e[i].f){
				dis[e[i].to]=dis[x]+e[i].v;
				from[e[i].to]=i;
				if (!vis[e[i].to]){
					vis[e[i].to]=1;
					q[++t]=e[i].to;
				}
			}
	}
	return dis[T]!=-(1ll<<60);
}
void del(){
	ans+=dis[T];
	for (int i=from[T];i;i=from[e[i].from])
		e[i].f--,e[i^1].f++;
}
int main(){
	scanf("%d",&n);
	nd=2*n; S=++nd; T=++nd;
	p0=++nd; p1=++nd; p2=++nd; p3=++nd;
	For(i,1,n){
		int x,y,v;
		scanf("%d%d%d",&x,&y,&v);
		add(S,i,v,0);
		add(i,p0,1<<30,x+y);
		add(i,p1,1<<30,x-y);
		add(i,p2,1<<30,-x+y);
		add(i,p3,1<<30,-x-y);
	}
	For(i,1,n){
		int x,y,v;
		scanf("%d%d%d",&x,&y,&v);
		add(i+n,T,v,0);
		add(p0,i+n,1<<30,-x-y);
		add(p1,i+n,1<<30,-x+y);
		add(p2,i+n,1<<30,x-y);
		add(p3,i+n,1<<30,x+y);
	}
	while (bfs()) del();
	printf("%lld\n",ans);
}
### 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>
#define ll long long
#define ull unsigned ll
#define uint unsigned
#define pii pair<int,int>
#define pll pair<ll,ll>
#define PB push_back
#define fi first
#define se second
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define CLR(a,v) memset(a,v,sizeof(a));
#define CPY(a,b) memcpy(a,b,sizeof(a));
using namespace std;
const int N=2010;
struct edge{
	int from,to,next,f,v;
}e[N*10];
int head[N],tot=1;
int n,nd,S,T,p0,p1,p2,p3;
int vis[N],q[N*10],from[N];
ll dis[N],ans;
void add(int x,int y,int f,int v){
	e[++tot]=(edge){x,y,head[x],f,v};
	head[x]=tot;
	e[++tot]=(edge){y,x,head[y],0,-v};
	head[y]=tot;
}
bool bfs(){
	For(i,1,nd){
		vis[i]=0;
		dis[i]=-(1ll<<60);
	}
	int h=0,t=1;
	q[1]=S; dis[S]=0;
	while (h!=t){
		int x=q[++h]; vis[x]=0;
		for (int i=head[x];i;i=e[i].next)
			if (dis[e[i].to]<dis[x]+e[i].v&&e[i].f){
				dis[e[i].to]=dis[x]+e[i].v;
				from[e[i].to]=i;
				if (!vis[e[i].to]){
					vis[e[i].to]=1;
					q[++t]=e[i].to;
				}
			}
	}
	return dis[T]!=-(1ll<<60);
}
void del(){
	ans+=dis[T];
	for (int i=from[T];i;i=from[e[i].from])
		e[i].f--,e[i^1].f++;
}
int main(){
	scanf("%d",&n);
	nd=2*n; S=++nd; T=++nd;
	p0=++nd; p1=++nd; p2=++nd; p3=++nd;
	For(i,1,n){
		int x,y,v;
		scanf("%d%d%d",&x,&y,&v);
		add(S,i,v,0);
		add(i,p0,1<<30,x+y);
		add(i,p1,1<<30,x-y);
		add(i,p2,1<<30,-x+y);
		add(i,p3,1<<30,-x-y);
	}
	For(i,1,n){
		int x,y,v;
		scanf("%d%d%d",&x,&y,&v);
		add(i+n,T,v,0);
		add(p0,i+n,1<<30,-x-y);
		add(p1,i+n,1<<30,-x+y);
		add(p2,i+n,1<<30,x-y);
		add(p3,i+n,1<<30,x+y);
	}
	while (bfs()) del();
	printf("%lld\n",ans);
}
```