#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
int S=0,T;
struct edge{
	int s,t,cap;long long v;int nxt;
}e[50005];
int e_cnt,last[5005];
void addedge(int u,int v,int cap,long long l){
	e[e_cnt]=(edge){u,v,cap,l,last[u]};
	last[u]=e_cnt++;
	e[e_cnt]=(edge){v,u,0,-l,last[v]};
	last[v]=e_cnt++;
}
long long d[5005];int fa[5005],Mn[5005];
bool in[5005];
int q[5005],h,t;
bool spfa(){
	for(int i=0;i<=T;i++)d[i]=-1e18;
	for(int i=0;i<=T;i++)Mn[i]=0;
	memset(in,0,sizeof(in));
	h=t=-1;
	d[S]=0; Mn[S]=1e9;
	q[++t]=S;
	while(h!=t){
		h=(h+1)%5000;
		int x=q[h];
		// cerr<<x<<endl;
		for(int i=last[x];i!=-1;i=e[i].nxt)if(e[i].cap&&d[e[i].t]<d[x]+e[i].v){
			d[e[i].t]=d[x]+e[i].v;
			fa[e[i].t]=i;
			Mn[e[i].t]=min(Mn[x],e[i].cap);
			if(!in[e[i].t])q[t=(t+1)%5000]=e[i].t,in[e[i].t]=1;
		}
		in[x]=0;
	}
	return Mn[T]>0;
}
pair<int,long long> solve(){
	int flow=0;long long cost=0;
	while(spfa()){
		// cerr<<flow<<endl;
		flow+=Mn[T];
		cost+=d[T]*Mn[T];
		int u=T;
		while(u!=S){
			e[fa[u]].cap-=Mn[T];
			e[fa[u]^1].cap+=Mn[T];
			u=e[fa[u]].s;
		}
	}
	return make_pair(flow,cost);
}
int x0[1005],y0[1005],c0[1005],x1[1005],y1[1005],c1[1005];
int main(){
	//freopen("a.in","r",stdin);
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d%d%d",&x0[i],&y0[i],&c0[i]);
	for(int i=1;i<=n;i++)
		scanf("%d%d%d",&x1[i],&y1[i],&c1[i]);
	memset(last,-1,sizeof(last));
	for(int i=1;i<=n;i++)addedge(S,i,c0[i],0);
	for(int i=1;i<=n;i++){
		addedge(i,n+1,1e9,x0[i]+y0[i]);
		addedge(i,n+2,1e9,x0[i]-y0[i]);
		addedge(i,n+3,1e9,-x0[i]+y0[i]);
		addedge(i,n+4,1e9,-x0[i]-y0[i]);
		
		addedge(n+1,i+n+4,1e9,-x1[i]-y1[i]);
		addedge(n+2,i+n+4,1e9,-x1[i]+y1[i]);
		addedge(n+3,i+n+4,1e9,x1[i]-y1[i]);
		addedge(n+4,i+n+4,1e9,x1[i]+y1[i]);
	}
	T=2*n+5;
	for(int i=1;i<=n;i++)addedge(i+n+4,T,c1[i],0);
	printf("%lld\n",solve().second);
	return 0;
}
### 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<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
int S=0,T;
struct edge{
	int s,t,cap;long long v;int nxt;
}e[50005];
int e_cnt,last[5005];
void addedge(int u,int v,int cap,long long l){
	e[e_cnt]=(edge){u,v,cap,l,last[u]};
	last[u]=e_cnt++;
	e[e_cnt]=(edge){v,u,0,-l,last[v]};
	last[v]=e_cnt++;
}
long long d[5005];int fa[5005],Mn[5005];
bool in[5005];
int q[5005],h,t;
bool spfa(){
	for(int i=0;i<=T;i++)d[i]=-1e18;
	for(int i=0;i<=T;i++)Mn[i]=0;
	memset(in,0,sizeof(in));
	h=t=-1;
	d[S]=0; Mn[S]=1e9;
	q[++t]=S;
	while(h!=t){
		h=(h+1)%5000;
		int x=q[h];
		// cerr<<x<<endl;
		for(int i=last[x];i!=-1;i=e[i].nxt)if(e[i].cap&&d[e[i].t]<d[x]+e[i].v){
			d[e[i].t]=d[x]+e[i].v;
			fa[e[i].t]=i;
			Mn[e[i].t]=min(Mn[x],e[i].cap);
			if(!in[e[i].t])q[t=(t+1)%5000]=e[i].t,in[e[i].t]=1;
		}
		in[x]=0;
	}
	return Mn[T]>0;
}
pair<int,long long> solve(){
	int flow=0;long long cost=0;
	while(spfa()){
		// cerr<<flow<<endl;
		flow+=Mn[T];
		cost+=d[T]*Mn[T];
		int u=T;
		while(u!=S){
			e[fa[u]].cap-=Mn[T];
			e[fa[u]^1].cap+=Mn[T];
			u=e[fa[u]].s;
		}
	}
	return make_pair(flow,cost);
}
int x0[1005],y0[1005],c0[1005],x1[1005],y1[1005],c1[1005];
int main(){
	//freopen("a.in","r",stdin);
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d%d%d",&x0[i],&y0[i],&c0[i]);
	for(int i=1;i<=n;i++)
		scanf("%d%d%d",&x1[i],&y1[i],&c1[i]);
	memset(last,-1,sizeof(last));
	for(int i=1;i<=n;i++)addedge(S,i,c0[i],0);
	for(int i=1;i<=n;i++){
		addedge(i,n+1,1e9,x0[i]+y0[i]);
		addedge(i,n+2,1e9,x0[i]-y0[i]);
		addedge(i,n+3,1e9,-x0[i]+y0[i]);
		addedge(i,n+4,1e9,-x0[i]-y0[i]);
		
		addedge(n+1,i+n+4,1e9,-x1[i]-y1[i]);
		addedge(n+2,i+n+4,1e9,-x1[i]+y1[i]);
		addedge(n+3,i+n+4,1e9,x1[i]-y1[i]);
		addedge(n+4,i+n+4,1e9,x1[i]+y1[i]);
	}
	T=2*n+5;
	for(int i=1;i<=n;i++)addedge(i+n+4,T,c1[i],0);
	printf("%lld\n",solve().second);
	return 0;
}
```