#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
typedef long long LL;
const int N = 2e4+10;
int to[N],cap[N],nxt[N],head[N],q[N<<5],v[N],path[N];
int n,x,y,c,l,r,i,tot=1,S,T;
LL cost[N], d[N], ans, inf=1e14;

void add(int x,int y,int z,int w){
	to[++tot]=y,cap[tot]=z,cost[tot]=w,nxt[tot]=head[x],head[x]=tot;
	to[++tot]=x,cap[tot]=0,cost[tot]=-w,nxt[tot]=head[y],head[y]=tot;
}

bool spfa(){
	rep(i,S,T)d[i]=inf;
	memset(v,0,sizeof(v));
	q[l=r=1]=S,d[S]=0,v[S]=1;
	while(l<=r){
		x=q[l++],v[x]=0;
		for(i=head[x];i;i=nxt[i])
			if(cap[i]&&d[y=to[i]]>d[x]+cost[i]){
				d[y]=d[x]+cost[i],path[y]=i;
				if(!v[y]) q[++r]=y,v[y]=1;
			}
	}
	return d[T]!=inf;
}

void update(){
	for(y=T;y!=S;y=x){
		i=path[y],x=to[i^1];
		--cap[i],++cap[i^1];
	}
	ans+=d[T];
}

int main(){
	ios::sync_with_stdio(false);cin.tie(0);
	cin>>n;
	S=0,T=n*2+5;
	rep(i,1,n){
		cin>>x>>y>>c;
		add(S,i,c,0);
		add(i,n+1,inf,-x-y);
		add(i,n+2,inf,-x+y);
		add(i,n+3,inf,x-y);
		add(i,n+4,inf,x+y);
	}
	rep(i,1,n){
		cin>>x>>y>>c;
		add(n+4+i,T,c,0);
		add(n+1,n+4+i,inf,x+y);
		add(n+2,n+4+i,inf,x-y);
		add(n+3,n+4+i,inf,-x+y);
		add(n+4,n+4+i,inf,-x-y);
	}
	while(spfa())update();
	cout<<(-ans)<<endl;
	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<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
typedef long long LL;
const int N = 2e4+10;
int to[N],cap[N],nxt[N],head[N],q[N<<5],v[N],path[N];
int n,x,y,c,l,r,i,tot=1,S,T;
LL cost[N], d[N], ans, inf=1e14;

void add(int x,int y,int z,int w){
	to[++tot]=y,cap[tot]=z,cost[tot]=w,nxt[tot]=head[x],head[x]=tot;
	to[++tot]=x,cap[tot]=0,cost[tot]=-w,nxt[tot]=head[y],head[y]=tot;
}

bool spfa(){
	rep(i,S,T)d[i]=inf;
	memset(v,0,sizeof(v));
	q[l=r=1]=S,d[S]=0,v[S]=1;
	while(l<=r){
		x=q[l++],v[x]=0;
		for(i=head[x];i;i=nxt[i])
			if(cap[i]&&d[y=to[i]]>d[x]+cost[i]){
				d[y]=d[x]+cost[i],path[y]=i;
				if(!v[y]) q[++r]=y,v[y]=1;
			}
	}
	return d[T]!=inf;
}

void update(){
	for(y=T;y!=S;y=x){
		i=path[y],x=to[i^1];
		--cap[i],++cap[i^1];
	}
	ans+=d[T];
}

int main(){
	ios::sync_with_stdio(false);cin.tie(0);
	cin>>n;
	S=0,T=n*2+5;
	rep(i,1,n){
		cin>>x>>y>>c;
		add(S,i,c,0);
		add(i,n+1,inf,-x-y);
		add(i,n+2,inf,-x+y);
		add(i,n+3,inf,x-y);
		add(i,n+4,inf,x+y);
	}
	rep(i,1,n){
		cin>>x>>y>>c;
		add(n+4+i,T,c,0);
		add(n+1,n+4+i,inf,x+y);
		add(n+2,n+4+i,inf,x-y);
		add(n+3,n+4+i,inf,-x+y);
		add(n+4,n+4+i,inf,-x-y);
	}
	while(spfa())update();
	cout<<(-ans)<<endl;
	return 0;
}
```