#include<cstdio>
#include<cstring>
#define inf 1000000000
using namespace std;
typedef long long ll;
const int N=2e3+100;
struct edge{
	int x,y,c1,c2,next;
}a[20*N];int len,last[N];
int st,ed,maxn;
void ins(int x,int y,int c1,int c2){
	a[++len].x=x;a[len].y=y;a[len].c1=c1;a[len].c2=c2;
	a[len].next=last[x];last[x]=len;
	a[++len].x=y;a[len].y=x;a[len].c1=0;a[len].c2=-c2;
	a[len].next=last[y];last[y]=len;
}
int dx[4]={-1,1,-1,1};
int dy[4]={1,1,-1,-1};
int list[N],head,tail,fa[N],Min[N];bool v[N];ll dis[N];
int mymin(int x,int y) {return x<y?x:y;}
int main()
{
	int n;scanf("%d",&n);
	maxn=2*n+6;st=maxn-1;ed=maxn;
	len=1;memset(last,0,sizeof(last));
	for(int i=1;i<=n;i++){
		int x,y,c;
		scanf("%d%d%d",&x,&y,&c);
		ins(st,i,c,0);
		for(int j=0;j<=3;j++){
			int xx=dx[j]*x,yy=dy[j]*y;
			ins(i,2*n+j+1,inf,xx+yy);
		}
	}
	for(int i=1;i<=n;i++){
		int x,y,c;
		scanf("%d%d%d",&x,&y,&c);
		ins(i+n,ed,c,0);
		for(int j=0;j<=3;j++){
			int xx=-dx[j]*x,yy=-dy[j]*y;
			ins(2*n+j+1,i+n,inf,xx+yy);
		}
	}
	ll ans=0;
	while(1){
		Min[st]=inf;
		memset(dis,-60,sizeof(dis));dis[st]=0;
		memset(v,true,sizeof(v));v[st]=false;
		list[1]=st;head=1;tail=2;if(tail==maxn+1) tail=1;
		while(head!=tail){
			int x=list[head];
			for(int k=last[x];k;k=a[k].next){
				int y=a[k].y;
				if(a[k].c1&&dis[y]<dis[x]+a[k].c2){
					fa[y]=k;
					dis[y]=dis[x]+a[k].c2;
					Min[y]=mymin(Min[x],a[k].c1);
					if(v[y]){
						v[y]=false;
						list[tail]=y;
						tail++;if(tail==maxn+1) tail=1;
					}
				}
			}
			head++;if(head==maxn+1) head=1;
			v[x]=true;
		}
		if(dis[ed]<0) break;
		ans=ans+dis[ed]*Min[ed];
		int t=ed;
		while(t!=st) a[fa[t]].c1-=Min[ed],a[fa[t]^1].c1+=Min[ed],t=a[fa[t]].x;
	}
	printf("%lld\n",ans);
	return 0;
}
### Prompt

Create a solution in cpp for 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>
#define inf 1000000000
using namespace std;
typedef long long ll;
const int N=2e3+100;
struct edge{
	int x,y,c1,c2,next;
}a[20*N];int len,last[N];
int st,ed,maxn;
void ins(int x,int y,int c1,int c2){
	a[++len].x=x;a[len].y=y;a[len].c1=c1;a[len].c2=c2;
	a[len].next=last[x];last[x]=len;
	a[++len].x=y;a[len].y=x;a[len].c1=0;a[len].c2=-c2;
	a[len].next=last[y];last[y]=len;
}
int dx[4]={-1,1,-1,1};
int dy[4]={1,1,-1,-1};
int list[N],head,tail,fa[N],Min[N];bool v[N];ll dis[N];
int mymin(int x,int y) {return x<y?x:y;}
int main()
{
	int n;scanf("%d",&n);
	maxn=2*n+6;st=maxn-1;ed=maxn;
	len=1;memset(last,0,sizeof(last));
	for(int i=1;i<=n;i++){
		int x,y,c;
		scanf("%d%d%d",&x,&y,&c);
		ins(st,i,c,0);
		for(int j=0;j<=3;j++){
			int xx=dx[j]*x,yy=dy[j]*y;
			ins(i,2*n+j+1,inf,xx+yy);
		}
	}
	for(int i=1;i<=n;i++){
		int x,y,c;
		scanf("%d%d%d",&x,&y,&c);
		ins(i+n,ed,c,0);
		for(int j=0;j<=3;j++){
			int xx=-dx[j]*x,yy=-dy[j]*y;
			ins(2*n+j+1,i+n,inf,xx+yy);
		}
	}
	ll ans=0;
	while(1){
		Min[st]=inf;
		memset(dis,-60,sizeof(dis));dis[st]=0;
		memset(v,true,sizeof(v));v[st]=false;
		list[1]=st;head=1;tail=2;if(tail==maxn+1) tail=1;
		while(head!=tail){
			int x=list[head];
			for(int k=last[x];k;k=a[k].next){
				int y=a[k].y;
				if(a[k].c1&&dis[y]<dis[x]+a[k].c2){
					fa[y]=k;
					dis[y]=dis[x]+a[k].c2;
					Min[y]=mymin(Min[x],a[k].c1);
					if(v[y]){
						v[y]=false;
						list[tail]=y;
						tail++;if(tail==maxn+1) tail=1;
					}
				}
			}
			head++;if(head==maxn+1) head=1;
			v[x]=true;
		}
		if(dis[ed]<0) break;
		ans=ans+dis[ed]*Min[ed];
		int t=ed;
		while(t!=st) a[fa[t]].c1-=Min[ed],a[fa[t]^1].c1+=Min[ed],t=a[fa[t]].x;
	}
	printf("%lld\n",ans);
	return 0;
}
```