#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll inf=1e15;
const int MAXN=2e3+10;
const int MAXM=4e4+20;
int edge=1,head[MAXN],tail[MAXM],nex[MAXM];
ll dist[MAXN],r[MAXM],w[MAXM];
int n,S,T,pree[MAXN],preu[MAXN];
int q[MAXM];
bool use[MAXN];
ll flow=0,cost=0;
// |x1-x2|+|y1-y2|=max(x1+y1-(x2+y2),x1-y1-(x2-y2),(y1-x1)-(y2-x2),-(x1+y1)+x2+y2
void add(int u,int v,ll R,ll W){
	edge++,nex[edge]=head[u],head[u]=edge,tail[edge]=v,r[edge]=R,w[edge]=W;
}
void ins(int u,int v,ll R,ll W){
	add(u,v,R,-W); add(v,u,0,W);
}
bool spfa(){
	for (int i=1;i<=T;i++) use[i]=0,dist[i]=inf;
	dist[S]=0; int L=0,R=0; q[++R]=S; use[S]=1;
	while (L<R){
		int u=q[++L];
		for (int e=head[u];e;e=nex[e]){
			int v=tail[e];
			if (dist[v]>dist[u]+w[e]&&r[e]>0){
				dist[v]=dist[u]+w[e];
				preu[v]=u; pree[v]=e;
				if (!use[v]) q[++R]=v,use[v]=1;
			}
		}
		use[u]=0;
	}
	return dist[T]<inf;
}
void argument(){
	ll res=inf;
	for (int now=T;now!=S;now=preu[now]){
		int e=pree[now];
		res=min(res,r[e]);
	}
	flow+=res; cost+=res*dist[T];
	for (int now=T;now!=S;now=preu[now]){
		int e=pree[now];
		r[e]-=res; r[e^1]+=res;
	}
}
int main(){
	scanf("%d",&n);
	S=2*n+5; T=S+1;
	for (int i=1;i<=n;i++){
		int x,y,c; scanf("%d%d%d",&x,&y,&c);
		ins(S,i,c,0);
		ins(i,2*n+1,inf,x+y);
		ins(i,2*n+2,inf,x-y);
		ins(i,2*n+3,inf,y-x);
		ins(i,2*n+4,inf,-x-y);
	}
	for (int i=1;i<=n;i++){
		int x,y,c; scanf("%d%d%d",&x,&y,&c);
		ins(i+n,T,c,0);
		ins(2*n+1,i+n,inf,-x-y);
		ins(2*n+2,i+n,inf,y-x);
		ins(2*n+3,i+n,inf,x-y);
		ins(2*n+4,i+n,inf,x+y);
	}
	while (spfa()) argument();
	printf("%lld\n",-cost);
	return 0;
}

### Prompt

In cpp, your task is to solve 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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll inf=1e15;
const int MAXN=2e3+10;
const int MAXM=4e4+20;
int edge=1,head[MAXN],tail[MAXM],nex[MAXM];
ll dist[MAXN],r[MAXM],w[MAXM];
int n,S,T,pree[MAXN],preu[MAXN];
int q[MAXM];
bool use[MAXN];
ll flow=0,cost=0;
// |x1-x2|+|y1-y2|=max(x1+y1-(x2+y2),x1-y1-(x2-y2),(y1-x1)-(y2-x2),-(x1+y1)+x2+y2
void add(int u,int v,ll R,ll W){
	edge++,nex[edge]=head[u],head[u]=edge,tail[edge]=v,r[edge]=R,w[edge]=W;
}
void ins(int u,int v,ll R,ll W){
	add(u,v,R,-W); add(v,u,0,W);
}
bool spfa(){
	for (int i=1;i<=T;i++) use[i]=0,dist[i]=inf;
	dist[S]=0; int L=0,R=0; q[++R]=S; use[S]=1;
	while (L<R){
		int u=q[++L];
		for (int e=head[u];e;e=nex[e]){
			int v=tail[e];
			if (dist[v]>dist[u]+w[e]&&r[e]>0){
				dist[v]=dist[u]+w[e];
				preu[v]=u; pree[v]=e;
				if (!use[v]) q[++R]=v,use[v]=1;
			}
		}
		use[u]=0;
	}
	return dist[T]<inf;
}
void argument(){
	ll res=inf;
	for (int now=T;now!=S;now=preu[now]){
		int e=pree[now];
		res=min(res,r[e]);
	}
	flow+=res; cost+=res*dist[T];
	for (int now=T;now!=S;now=preu[now]){
		int e=pree[now];
		r[e]-=res; r[e^1]+=res;
	}
}
int main(){
	scanf("%d",&n);
	S=2*n+5; T=S+1;
	for (int i=1;i<=n;i++){
		int x,y,c; scanf("%d%d%d",&x,&y,&c);
		ins(S,i,c,0);
		ins(i,2*n+1,inf,x+y);
		ins(i,2*n+2,inf,x-y);
		ins(i,2*n+3,inf,y-x);
		ins(i,2*n+4,inf,-x-y);
	}
	for (int i=1;i<=n;i++){
		int x,y,c; scanf("%d%d%d",&x,&y,&c);
		ins(i+n,T,c,0);
		ins(2*n+1,i+n,inf,-x-y);
		ins(2*n+2,i+n,inf,y-x);
		ins(2*n+3,i+n,inf,x-y);
		ins(2*n+4,i+n,inf,x+y);
	}
	while (spfa()) argument();
	printf("%lld\n",-cost);
	return 0;
}

```