#include<bits/stdc++.h>
#define ll long long
#define inf 1000000007
using namespace std;
ll maxcost;
int n,S,T,cnt=-1;
int rx[1005],ry[1005],bx[1005],by[1005];
int numr[1005],numb[1005];
int fir[2105],nxt[3000005],to[3000005],flow[3000005];
ll dis[3000005],Dis[2105];
int Flow[2105],vis[2105],pre[2105],last[2105];

void add(int u,int v,int F,ll D){
	to[++cnt]=v,nxt[cnt]=fir[u],fir[u]=cnt;
	dis[cnt]=D,flow[cnt]=F;
}

int spfa(){
	memset(vis,0,sizeof(vis));
	for(int i=S;i<=T;i++) Dis[i]=-1e18,Flow[i]=inf;
	queue<int> q;
	q.push(S),Dis[S]=0,vis[S]=1,pre[T]=-1;
	while(!q.empty()){
		int now=q.front();
		q.pop();
		vis[now]=0;
		for(int i=fir[now];i!=-1;i=nxt[i]){
			if(flow[i]>0&&Dis[to[i]]<Dis[now]+dis[i]){
				Dis[to[i]]=Dis[now]+dis[i];
				pre[to[i]]=now;last[to[i]]=i;
				Flow[to[i]]=min(Flow[now],flow[i]);
				if(!vis[to[i]]) vis[to[i]]=1,q.push(to[i]);
			}
		}
	}
	return pre[T]!=-1;
}

void MCMF(){
	while(spfa()){
		int now=T;
		maxcost+=1ll*Flow[T]*Dis[T];
		while(now!=S){
			flow[last[now]]-=Flow[T];
			flow[last[now]^1]+=Flow[T];
			now=pre[now];
		}
	}
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d %d %d",&rx[i],&ry[i],&numr[i]);
	for(int i=1;i<=n;i++) scanf("%d %d %d",&bx[i],&by[i],&numb[i]);
	S=0,T=2*n+9;
	memset(fir,-1,sizeof(fir));
	for(int i=1;i<=n;i++) add(S,i,numr[i],0),add(i,S,0,0);
	for(int i=1;i<=n;i++) add(n+i,T,numb[i],0),add(T,n+i,0,0);
	for(int i=1;i<=n;i++){
		add(i,2*n+1,inf,-rx[i]-ry[i]),add(2*n+1,i,0,rx[i]+ry[i]);
		add(i,2*n+2,inf,-rx[i]+ry[i]),add(2*n+2,i,0,rx[i]-ry[i]);
		add(i,2*n+3,inf,rx[i]-ry[i]),add(2*n+3,i,0,-rx[i]+ry[i]);
		add(i,2*n+4,inf,rx[i]+ry[i]),add(2*n+4,i,0,-rx[i]-ry[i]);
	}
	for(int i=1;i<=n;i++){
		add(2*n+5,n+i,inf,bx[i]+by[i]),add(n+i,2*n+5,0,-bx[i]-by[i]);
		add(2*n+6,n+i,inf,bx[i]-by[i]),add(n+i,2*n+6,0,-bx[i]+by[i]);
		add(2*n+7,n+i,inf,-bx[i]+by[i]),add(n+i,2*n+7,0,bx[i]-by[i]);
		add(2*n+8,n+i,inf,-bx[i]-by[i]),add(n+i,2*n+8,0,bx[i]+by[i]);
	}
	for(int i=1;i<=4;i++) add(2*n+i,2*n+i+4,inf,0),add(2*n+i+4,2*n+i,0,0);
	MCMF();
	printf("%lld\n",maxcost);
	
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 inf 1000000007
using namespace std;
ll maxcost;
int n,S,T,cnt=-1;
int rx[1005],ry[1005],bx[1005],by[1005];
int numr[1005],numb[1005];
int fir[2105],nxt[3000005],to[3000005],flow[3000005];
ll dis[3000005],Dis[2105];
int Flow[2105],vis[2105],pre[2105],last[2105];

void add(int u,int v,int F,ll D){
	to[++cnt]=v,nxt[cnt]=fir[u],fir[u]=cnt;
	dis[cnt]=D,flow[cnt]=F;
}

int spfa(){
	memset(vis,0,sizeof(vis));
	for(int i=S;i<=T;i++) Dis[i]=-1e18,Flow[i]=inf;
	queue<int> q;
	q.push(S),Dis[S]=0,vis[S]=1,pre[T]=-1;
	while(!q.empty()){
		int now=q.front();
		q.pop();
		vis[now]=0;
		for(int i=fir[now];i!=-1;i=nxt[i]){
			if(flow[i]>0&&Dis[to[i]]<Dis[now]+dis[i]){
				Dis[to[i]]=Dis[now]+dis[i];
				pre[to[i]]=now;last[to[i]]=i;
				Flow[to[i]]=min(Flow[now],flow[i]);
				if(!vis[to[i]]) vis[to[i]]=1,q.push(to[i]);
			}
		}
	}
	return pre[T]!=-1;
}

void MCMF(){
	while(spfa()){
		int now=T;
		maxcost+=1ll*Flow[T]*Dis[T];
		while(now!=S){
			flow[last[now]]-=Flow[T];
			flow[last[now]^1]+=Flow[T];
			now=pre[now];
		}
	}
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d %d %d",&rx[i],&ry[i],&numr[i]);
	for(int i=1;i<=n;i++) scanf("%d %d %d",&bx[i],&by[i],&numb[i]);
	S=0,T=2*n+9;
	memset(fir,-1,sizeof(fir));
	for(int i=1;i<=n;i++) add(S,i,numr[i],0),add(i,S,0,0);
	for(int i=1;i<=n;i++) add(n+i,T,numb[i],0),add(T,n+i,0,0);
	for(int i=1;i<=n;i++){
		add(i,2*n+1,inf,-rx[i]-ry[i]),add(2*n+1,i,0,rx[i]+ry[i]);
		add(i,2*n+2,inf,-rx[i]+ry[i]),add(2*n+2,i,0,rx[i]-ry[i]);
		add(i,2*n+3,inf,rx[i]-ry[i]),add(2*n+3,i,0,-rx[i]+ry[i]);
		add(i,2*n+4,inf,rx[i]+ry[i]),add(2*n+4,i,0,-rx[i]-ry[i]);
	}
	for(int i=1;i<=n;i++){
		add(2*n+5,n+i,inf,bx[i]+by[i]),add(n+i,2*n+5,0,-bx[i]-by[i]);
		add(2*n+6,n+i,inf,bx[i]-by[i]),add(n+i,2*n+6,0,-bx[i]+by[i]);
		add(2*n+7,n+i,inf,-bx[i]+by[i]),add(n+i,2*n+7,0,bx[i]-by[i]);
		add(2*n+8,n+i,inf,-bx[i]-by[i]),add(n+i,2*n+8,0,bx[i]+by[i]);
	}
	for(int i=1;i<=4;i++) add(2*n+i,2*n+i+4,inf,0),add(2*n+i+4,2*n+i,0,0);
	MCMF();
	printf("%lld\n",maxcost);
	
	return 0;
}
```