#include<iostream>
#include<cstdio>
#include<queue>
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define REP(u) for(int i=hd[u],v,w,c;v=e[i].v,w=e[i].w,c=e[i].c,i;i=e[i].n)
using namespace std;
typedef long long ll;
const int N=100100,M=N<<1;
struct edge{int n,v,w,c;}e[M];
int n,T,fl=1,x,y,c,hd[N],p[N],I[N],f[N],in[N];
ll ans,d[N];
queue<int>q;
void add_(int u,int v,int w,int c){e[++fl]={hd[u],v,w,c};hd[u]=fl;}
void adde(int u,int v,int w,int c){add_(u,v,w,c);add_(v,u,-w,0);}
bool SPFA(){
	FOR(i,2,T) d[i]=-1e15;
	q.push(1);f[1]=1e9;
	while(q.size()){
		int u=q.front();q.pop();in[u]=0;
		REP(u)if(c && d[v]<d[u]+w){
			d[v]=d[u]+w,f[v]=min(f[u],c),p[v]=u,I[v]=i;
			if(!in[v]) in[v]=1,q.push(v);
		}
	}return d[T]>=0;
}
int main(){
	scanf("%d",&n);T=2*n+6;
	FOR(i,1,n){
		scanf("%d%d%d",&x,&y,&c);
		adde(1,i+5,0,c);
		adde(i+5,2,x+y,c);adde(i+5,3,y-x,c);
		adde(i+5,4,x-y,c);adde(i+5,5,-x-y,c);
	}
	FOR(i,n+1,n*2){
		scanf("%d%d%d",&x,&y,&c);
		adde(i+5,T,0,c);
		adde(2,i+5,-x-y,c);adde(3,i+5,x-y,c);
		adde(4,i+5,y-x,c); adde(5,i+5,x+y,c);
	}
	while(SPFA())for(int u=T;u!=1;u=p[u])
		e[I[u]].c-=f[T],e[I[u]^1].c+=f[T],ans+=1ll*e[I[u]].w*f[T];
	cout<<ans<<'\n';
}
### Prompt

Your challenge is to write 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<iostream>
#include<cstdio>
#include<queue>
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define REP(u) for(int i=hd[u],v,w,c;v=e[i].v,w=e[i].w,c=e[i].c,i;i=e[i].n)
using namespace std;
typedef long long ll;
const int N=100100,M=N<<1;
struct edge{int n,v,w,c;}e[M];
int n,T,fl=1,x,y,c,hd[N],p[N],I[N],f[N],in[N];
ll ans,d[N];
queue<int>q;
void add_(int u,int v,int w,int c){e[++fl]={hd[u],v,w,c};hd[u]=fl;}
void adde(int u,int v,int w,int c){add_(u,v,w,c);add_(v,u,-w,0);}
bool SPFA(){
	FOR(i,2,T) d[i]=-1e15;
	q.push(1);f[1]=1e9;
	while(q.size()){
		int u=q.front();q.pop();in[u]=0;
		REP(u)if(c && d[v]<d[u]+w){
			d[v]=d[u]+w,f[v]=min(f[u],c),p[v]=u,I[v]=i;
			if(!in[v]) in[v]=1,q.push(v);
		}
	}return d[T]>=0;
}
int main(){
	scanf("%d",&n);T=2*n+6;
	FOR(i,1,n){
		scanf("%d%d%d",&x,&y,&c);
		adde(1,i+5,0,c);
		adde(i+5,2,x+y,c);adde(i+5,3,y-x,c);
		adde(i+5,4,x-y,c);adde(i+5,5,-x-y,c);
	}
	FOR(i,n+1,n*2){
		scanf("%d%d%d",&x,&y,&c);
		adde(i+5,T,0,c);
		adde(2,i+5,-x-y,c);adde(3,i+5,x-y,c);
		adde(4,i+5,y-x,c); adde(5,i+5,x+y,c);
	}
	while(SPFA())for(int u=T;u!=1;u=p[u])
		e[I[u]].c-=f[T],e[I[u]^1].c+=f[T],ans+=1ll*e[I[u]].w*f[T];
	cout<<ans<<'\n';
}
```