#include<bits/stdc++.h>
#define fo(i,a,b)for(int i=a,_e=b;i<=_e;++i)
#define fd(i,a,b)for(int i=b,_e=a;i>=_e;--i)
#define ll long long
using namespace std;
const int N=1005;
int n,x,y,w,i,js,la[N*2],T;
int d[N*N*2],fr[N*2];
ll ans,f[N*2];
bool vi[N*2];
struct edge{
	int ne,y,l,c;
}e[N*20];
void link(int x,int y,int l,int c){
	e[++js]=(edge){la[x],y,l,c};la[x]=js;
	e[++js]=(edge){la[y],x,0,-c};la[y]=js;
}
int main(){
	cin>>n;js=1;T=n+n+5;
	fo(i,1,n){
		cin>>x>>y>>w;
		link(0,i,w,0);
		link(i,n+n+1,w,-x-y);
		link(i,n+n+2,w,-x+y);
		link(i,n+n+3,w,x-y);
		link(i,n+n+4,w,x+y);
	}
	fo(i,1,n){
		cin>>x>>y>>w;
		link(n+i,T,w,0);
		link(n+n+1,n+i,w,x+y);
		link(n+n+2,n+i,w,x-y);
		link(n+n+3,n+i,w,-x+y);
		link(n+n+4,n+i,w,-x-y);
	}
	for(;;){
		fo(i,0,T)f[i]=1e18;
		f[0]=0;vi[0]=1;
		for(int l=0,r=1,x,y;x=d[++l],l<=r;vi[x]=0)
			for(int i=la[x];i;i=e[i].ne)
				if(e[i].l&&f[x]+e[i].c<f[y=e[i].y]){
					f[y]=f[x]+e[i].c;
					fr[y]=i^1;
					if(!vi[y])d[++r]=y,vi[y]=1;
				}
		if(f[T]==1e18)break;
		int mi=10000;
		for(int x=T;x;x=e[fr[x]].y)
			mi=min(mi,e[fr[x]^1].l);
		ans+=f[T]*mi;
		for(int x=T;x;x=e[fr[x]].y)
			e[fr[x]].l+=mi,e[fr[x]^1].l-=mi;
	}
	cout<<-ans;
}
### Prompt

Your task is to create 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>
#define fo(i,a,b)for(int i=a,_e=b;i<=_e;++i)
#define fd(i,a,b)for(int i=b,_e=a;i>=_e;--i)
#define ll long long
using namespace std;
const int N=1005;
int n,x,y,w,i,js,la[N*2],T;
int d[N*N*2],fr[N*2];
ll ans,f[N*2];
bool vi[N*2];
struct edge{
	int ne,y,l,c;
}e[N*20];
void link(int x,int y,int l,int c){
	e[++js]=(edge){la[x],y,l,c};la[x]=js;
	e[++js]=(edge){la[y],x,0,-c};la[y]=js;
}
int main(){
	cin>>n;js=1;T=n+n+5;
	fo(i,1,n){
		cin>>x>>y>>w;
		link(0,i,w,0);
		link(i,n+n+1,w,-x-y);
		link(i,n+n+2,w,-x+y);
		link(i,n+n+3,w,x-y);
		link(i,n+n+4,w,x+y);
	}
	fo(i,1,n){
		cin>>x>>y>>w;
		link(n+i,T,w,0);
		link(n+n+1,n+i,w,x+y);
		link(n+n+2,n+i,w,x-y);
		link(n+n+3,n+i,w,-x+y);
		link(n+n+4,n+i,w,-x-y);
	}
	for(;;){
		fo(i,0,T)f[i]=1e18;
		f[0]=0;vi[0]=1;
		for(int l=0,r=1,x,y;x=d[++l],l<=r;vi[x]=0)
			for(int i=la[x];i;i=e[i].ne)
				if(e[i].l&&f[x]+e[i].c<f[y=e[i].y]){
					f[y]=f[x]+e[i].c;
					fr[y]=i^1;
					if(!vi[y])d[++r]=y,vi[y]=1;
				}
		if(f[T]==1e18)break;
		int mi=10000;
		for(int x=T;x;x=e[fr[x]].y)
			mi=min(mi,e[fr[x]^1].l);
		ans+=f[T]*mi;
		for(int x=T;x;x=e[fr[x]].y)
			e[fr[x]].l+=mi,e[fr[x]^1].l-=mi;
	}
	cout<<-ans;
}
```