#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<queue>
#include<cmath>
#include<cstdlib>
#define LL long long
#define LD long double
using namespace std;
const int NN=1000 +117;
const int MM=1000 +117;
int read(){
	int fl=1,x;char c;
	for(c=getchar();(c<'0'||c>'9')&&c!='-';c=getchar());
	if(c=='-'){fl=-1;c=getchar();}
	for(x=0;c>='0'&&c<='9';c=getchar())
		x=(x<<3)+(x<<1)+c-'0';
	return x*fl;
}
void open(){
	freopen("a.in","r",stdin);
	//freopen("a.out","w",stdout);
}
void close(){
	fclose(stdin);
	fclose(stdout);
}


int m,n;
struct edge{
	int to,xt,rem;
	LL ct;
}e[100*NN];
int ecnt=2;
int p[2*NN]={};
void ADD(int a,int b,int cap,LL ct){
	e[ecnt].to=b;
	e[ecnt].xt=p[a];
	e[ecnt].rem=cap;
	e[ecnt].ct=ct;
	p[a]=ecnt++;
}
void add(int a,int b,int cap,LL ct){
	ADD(a,b,cap,ct);
	ADD(b,a,0,-ct);
}

int con[4][2]={{1,1},{1,-1},{-1,1},{-1,-1}};
struct node{
	LL x,y;
	LL v(int j){
		return x*con[j][0]+y*con[j][1];
	}
};

int src,fin;
int mid[4]={};
LL ans,flow;
int pre[2*NN]={};
LL sp[2*NN]={};
int pcnt=0;
queue<int> q;
bool inq[2*NN]={};
bool spfa(){
	for(int i=1;i<=pcnt;++i)sp[i]=1e18;
	sp[src]=0;
	q.push(src);
	while(!q.empty()){
		int x=q.front();
		q.pop();
		inq[x]=0;
		for(int pos=p[x];pos;pos=e[pos].xt){
			int cur=e[pos].to;
			if(e[pos].rem&&sp[cur]>e[pos].ct+sp[x]){
				sp[cur]=e[pos].ct+sp[x];
				pre[cur]=pos;
				if(!inq[cur]){
					inq[cur]=1;
					q.push(cur);
				}
			}
		}
	}
	return sp[fin]!=1e18;
}
void getf(){
	while(spfa()){
		int cap=1e9;
		int pos=fin;
		while(pos!=src){
			int x=pre[pos];
			cap=min(e[x].rem,cap);
			pos=e[x^1].to;
		}
		flow+=cap;
		ans+=cap*sp[fin];
		pos=fin;
		while(pos!=src){
			int x=pre[pos];
			e[x].rem-=cap;
			e[x^1].rem+=cap;
			pos=e[x^1].to;
		}
	}
}
int main(){
	//open();
	n=read();
	for(int i=0;i<4;++i){
		mid[i]=2*n+i+1;
	}
	src=2*n+5;
	fin=src+1;
	pcnt=fin;
	for(int i=1;i<=n;++i){
		node a;
		a.x=read(),a.y=read();
		int c=read();
		add(src,i,c,0);
		for(int j=0;j<4;++j){
			add(i,mid[j],1e9,a.v(j));
		}
	}
	for(int i=1;i<=n;++i){
		node a;
		a.x=read(),a.y=read();
		int c=read();
		add(i+n,fin,c,0);
		for(int j=0;j<4;++j){
			add(mid[j],i+n,1e9,a.v(3^j));
		}
	}
	getf();
	printf("%lld\n",-ans);

	close();
	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<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<queue>
#include<cmath>
#include<cstdlib>
#define LL long long
#define LD long double
using namespace std;
const int NN=1000 +117;
const int MM=1000 +117;
int read(){
	int fl=1,x;char c;
	for(c=getchar();(c<'0'||c>'9')&&c!='-';c=getchar());
	if(c=='-'){fl=-1;c=getchar();}
	for(x=0;c>='0'&&c<='9';c=getchar())
		x=(x<<3)+(x<<1)+c-'0';
	return x*fl;
}
void open(){
	freopen("a.in","r",stdin);
	//freopen("a.out","w",stdout);
}
void close(){
	fclose(stdin);
	fclose(stdout);
}


int m,n;
struct edge{
	int to,xt,rem;
	LL ct;
}e[100*NN];
int ecnt=2;
int p[2*NN]={};
void ADD(int a,int b,int cap,LL ct){
	e[ecnt].to=b;
	e[ecnt].xt=p[a];
	e[ecnt].rem=cap;
	e[ecnt].ct=ct;
	p[a]=ecnt++;
}
void add(int a,int b,int cap,LL ct){
	ADD(a,b,cap,ct);
	ADD(b,a,0,-ct);
}

int con[4][2]={{1,1},{1,-1},{-1,1},{-1,-1}};
struct node{
	LL x,y;
	LL v(int j){
		return x*con[j][0]+y*con[j][1];
	}
};

int src,fin;
int mid[4]={};
LL ans,flow;
int pre[2*NN]={};
LL sp[2*NN]={};
int pcnt=0;
queue<int> q;
bool inq[2*NN]={};
bool spfa(){
	for(int i=1;i<=pcnt;++i)sp[i]=1e18;
	sp[src]=0;
	q.push(src);
	while(!q.empty()){
		int x=q.front();
		q.pop();
		inq[x]=0;
		for(int pos=p[x];pos;pos=e[pos].xt){
			int cur=e[pos].to;
			if(e[pos].rem&&sp[cur]>e[pos].ct+sp[x]){
				sp[cur]=e[pos].ct+sp[x];
				pre[cur]=pos;
				if(!inq[cur]){
					inq[cur]=1;
					q.push(cur);
				}
			}
		}
	}
	return sp[fin]!=1e18;
}
void getf(){
	while(spfa()){
		int cap=1e9;
		int pos=fin;
		while(pos!=src){
			int x=pre[pos];
			cap=min(e[x].rem,cap);
			pos=e[x^1].to;
		}
		flow+=cap;
		ans+=cap*sp[fin];
		pos=fin;
		while(pos!=src){
			int x=pre[pos];
			e[x].rem-=cap;
			e[x^1].rem+=cap;
			pos=e[x^1].to;
		}
	}
}
int main(){
	//open();
	n=read();
	for(int i=0;i<4;++i){
		mid[i]=2*n+i+1;
	}
	src=2*n+5;
	fin=src+1;
	pcnt=fin;
	for(int i=1;i<=n;++i){
		node a;
		a.x=read(),a.y=read();
		int c=read();
		add(src,i,c,0);
		for(int j=0;j<4;++j){
			add(i,mid[j],1e9,a.v(j));
		}
	}
	for(int i=1;i<=n;++i){
		node a;
		a.x=read(),a.y=read();
		int c=read();
		add(i+n,fin,c,0);
		for(int j=0;j<4;++j){
			add(mid[j],i+n,1e9,a.v(3^j));
		}
	}
	getf();
	printf("%lld\n",-ans);

	close();
	return 0;
}
```