#include<cstdio>
#include<cstring>
#include<algorithm>
#define clr(a,x) memset(a,x,sizeof(a))
using namespace std;

typedef long long ll;
const int N=100007,INF=0x3f3f3f3f;
const ll MAX=(1ll<<32);
inline ll read(){
	ll x=0,f=0;
	char c=getchar();
	for(;c<'0'||c>'9';c=getchar())if(c=='-')f=1;
	for(;c>='0'&&c<='9';c=getchar())x=x*10+c-'0';
	return f?-x:x;
}

int tot=1,S,T,h,t,st[N],to[N],nx[N],len[N],maxflow[N],vis[N],q[N],pre[N];
ll ans,cost[N],dis[N];
void add(int u,int v,int w,ll c){
	c=MAX-c;
	to[++tot]=v,nx[tot]=st[u],st[u]=tot,len[tot]=w,cost[tot]=c;
	to[++tot]=u,nx[tot]=st[v],st[v]=tot,len[tot]=0,cost[tot]=-c;
}
int bfs(){
	clr(maxflow,0);
	clr(pre,0);
	clr(vis,0);
	clr(dis,0x3f);
	h=1,q[t=1]=S,maxflow[S]=INF,dis[S]=0,vis[S]=1;
	while(h<=t){
		int u=q[h++];
		vis[u]=0;
		for(int i=st[u],v;i;i=nx[i]){
			v=to[i];
			if(len[i]>0&&dis[u]+cost[i]<dis[v]){
				maxflow[v]=min(maxflow[u],len[i]),dis[v]=dis[u]+cost[i],pre[v]=i;
				if(!vis[v])q[++t]=v,vis[v]=1;
			}
		}
	}
	return maxflow[T]>0&&dis[T]>0;
}
void upd(){
	int u=T;
	while(u!=S)len[pre[u]]-=maxflow[T],len[pre[u]^1]+=maxflow[T],u=to[pre[u]^1];
	ans+=dis[T]*maxflow[T];
}

int n,c[2][N];
ll sum,x[2][N],y[2][N];

int main(){
	n=read(),S=2*n+5,T=2*n+6;
	for(int i=1;i<=n;++i){
		ll x=read(),y=read();
		int c=read();
		sum+=c;
		add(S,i,c,MAX);
		add(i,2*n+1,c,x+y);
		add(i,2*n+2,c,x-y);
		add(i,2*n+3,c,y-x);
		add(i,2*n+4,c,-x-y);
	}
	for(int i=1;i<=n;++i){
		ll x=read(),y=read();
		int c=read();
		add(i+n,T,c,MAX);
		add(2*n+1,i+n,c,-x-y);
		add(2*n+2,i+n,c,y-x);
		add(2*n+3,i+n,c,x-y);
		add(2*n+4,i+n,c,x+y);
	}
	while(bfs())upd();
	printf("%lld\n",sum*MAX*2-ans);
	return 0;
}

### Prompt

Please formulate 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>
#define clr(a,x) memset(a,x,sizeof(a))
using namespace std;

typedef long long ll;
const int N=100007,INF=0x3f3f3f3f;
const ll MAX=(1ll<<32);
inline ll read(){
	ll x=0,f=0;
	char c=getchar();
	for(;c<'0'||c>'9';c=getchar())if(c=='-')f=1;
	for(;c>='0'&&c<='9';c=getchar())x=x*10+c-'0';
	return f?-x:x;
}

int tot=1,S,T,h,t,st[N],to[N],nx[N],len[N],maxflow[N],vis[N],q[N],pre[N];
ll ans,cost[N],dis[N];
void add(int u,int v,int w,ll c){
	c=MAX-c;
	to[++tot]=v,nx[tot]=st[u],st[u]=tot,len[tot]=w,cost[tot]=c;
	to[++tot]=u,nx[tot]=st[v],st[v]=tot,len[tot]=0,cost[tot]=-c;
}
int bfs(){
	clr(maxflow,0);
	clr(pre,0);
	clr(vis,0);
	clr(dis,0x3f);
	h=1,q[t=1]=S,maxflow[S]=INF,dis[S]=0,vis[S]=1;
	while(h<=t){
		int u=q[h++];
		vis[u]=0;
		for(int i=st[u],v;i;i=nx[i]){
			v=to[i];
			if(len[i]>0&&dis[u]+cost[i]<dis[v]){
				maxflow[v]=min(maxflow[u],len[i]),dis[v]=dis[u]+cost[i],pre[v]=i;
				if(!vis[v])q[++t]=v,vis[v]=1;
			}
		}
	}
	return maxflow[T]>0&&dis[T]>0;
}
void upd(){
	int u=T;
	while(u!=S)len[pre[u]]-=maxflow[T],len[pre[u]^1]+=maxflow[T],u=to[pre[u]^1];
	ans+=dis[T]*maxflow[T];
}

int n,c[2][N];
ll sum,x[2][N],y[2][N];

int main(){
	n=read(),S=2*n+5,T=2*n+6;
	for(int i=1;i<=n;++i){
		ll x=read(),y=read();
		int c=read();
		sum+=c;
		add(S,i,c,MAX);
		add(i,2*n+1,c,x+y);
		add(i,2*n+2,c,x-y);
		add(i,2*n+3,c,y-x);
		add(i,2*n+4,c,-x-y);
	}
	for(int i=1;i<=n;++i){
		ll x=read(),y=read();
		int c=read();
		add(i+n,T,c,MAX);
		add(2*n+1,i+n,c,-x-y);
		add(2*n+2,i+n,c,y-x);
		add(2*n+3,i+n,c,x-y);
		add(2*n+4,i+n,c,x+y);
	}
	while(bfs())upd();
	printf("%lld\n",sum*MAX*2-ans);
	return 0;
}

```