#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=2010,M=4e6+10;int n;
const ll inf=1e16;
struct path{int y,nxt;ll w,cost;}g[M];int h[N],tot=1;
void add(int x,int y,ll w,ll cost){
	g[++tot]=(path){y,h[x],w,cost};h[x]=tot;
	g[++tot]=(path){x,h[y],0,-cost};h[y]=tot;
}
namespace SPFA{
	int S,T,pre[N],q[M],ql,qr;bool vis[N];ll dis[N];
	ll spfa(){
		for(int i=0;i<=T;i++)dis[i]=-inf;
		q[ql=qr=1]=S;dis[S]=0;vis[S]=1;pre[S]=0;
		while(ql<=qr){
			int x=q[ql++];vis[x]=0;
			for(int i=h[x];i;i=g[i].nxt)if(g[i].w&&dis[g[i].y]<g[i].cost+dis[x]){
				pre[g[i].y]=i;dis[g[i].y]=g[i].cost+dis[x];
				if(!vis[g[i].y])q[++qr]=g[i].y,vis[g[i].y]=1;
			}
		}
		if(dis[T]==-1)return 0;
		ll s=0,w=inf;
		for(int p=T,i=pre[T];p!=S;p=g[i^1].y,i=pre[p]){
			w=min(w,g[i].w);s+=g[i].cost;
		}
		for(int p=T,i=pre[T];p!=S;p=g[i^1].y,i=pre[p]){
			g[i].w-=w;g[i^1].w+=w;
		}return w*s;
	}
	ll solve(){
		ll ans=0,ret;while(ret=spfa())
		ans+=ret;
		return ans;
	}
}
int r[N][3],b[N][3],S,T;
int main(){
	//freopen("a.in","r",stdin);
	scanf("%d",&n);T=SPFA::T=2*n+5;
	for(int i=1;i<=n;i++)for(int j=0;j<3;j++)scanf("%d",&r[i][j]);
	for(int i=1;i<=n;i++)for(int j=0;j<3;j++)scanf("%d",&b[i][j]);
	for(int i=1;i<=n;i++)add(S,i,r[i][2],0),add(i+n,T,b[i][2],0);
	for(int i=1;i<=n;i++){
		add(i,n*2+1,inf,r[i][0]+r[i][1]);
		add(i,n*2+2,inf,-r[i][0]+r[i][1]);
		add(i,n*2+3,inf,r[i][0]-r[i][1]);
		add(i,n*2+4,inf,-r[i][0]-r[i][1]);

		add(n*2+1,i+n,inf,-b[i][0]-b[i][1]);
		add(n*2+2,i+n,inf,b[i][0]-b[i][1]);
		add(n*2+3,i+n,inf,-b[i][0]+b[i][1]);
		add(n*2+4,i+n,inf,b[i][0]+b[i][1]);
	}
	printf("%lld\n",SPFA::solve());
}
### 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<cmath>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=2010,M=4e6+10;int n;
const ll inf=1e16;
struct path{int y,nxt;ll w,cost;}g[M];int h[N],tot=1;
void add(int x,int y,ll w,ll cost){
	g[++tot]=(path){y,h[x],w,cost};h[x]=tot;
	g[++tot]=(path){x,h[y],0,-cost};h[y]=tot;
}
namespace SPFA{
	int S,T,pre[N],q[M],ql,qr;bool vis[N];ll dis[N];
	ll spfa(){
		for(int i=0;i<=T;i++)dis[i]=-inf;
		q[ql=qr=1]=S;dis[S]=0;vis[S]=1;pre[S]=0;
		while(ql<=qr){
			int x=q[ql++];vis[x]=0;
			for(int i=h[x];i;i=g[i].nxt)if(g[i].w&&dis[g[i].y]<g[i].cost+dis[x]){
				pre[g[i].y]=i;dis[g[i].y]=g[i].cost+dis[x];
				if(!vis[g[i].y])q[++qr]=g[i].y,vis[g[i].y]=1;
			}
		}
		if(dis[T]==-1)return 0;
		ll s=0,w=inf;
		for(int p=T,i=pre[T];p!=S;p=g[i^1].y,i=pre[p]){
			w=min(w,g[i].w);s+=g[i].cost;
		}
		for(int p=T,i=pre[T];p!=S;p=g[i^1].y,i=pre[p]){
			g[i].w-=w;g[i^1].w+=w;
		}return w*s;
	}
	ll solve(){
		ll ans=0,ret;while(ret=spfa())
		ans+=ret;
		return ans;
	}
}
int r[N][3],b[N][3],S,T;
int main(){
	//freopen("a.in","r",stdin);
	scanf("%d",&n);T=SPFA::T=2*n+5;
	for(int i=1;i<=n;i++)for(int j=0;j<3;j++)scanf("%d",&r[i][j]);
	for(int i=1;i<=n;i++)for(int j=0;j<3;j++)scanf("%d",&b[i][j]);
	for(int i=1;i<=n;i++)add(S,i,r[i][2],0),add(i+n,T,b[i][2],0);
	for(int i=1;i<=n;i++){
		add(i,n*2+1,inf,r[i][0]+r[i][1]);
		add(i,n*2+2,inf,-r[i][0]+r[i][1]);
		add(i,n*2+3,inf,r[i][0]-r[i][1]);
		add(i,n*2+4,inf,-r[i][0]-r[i][1]);

		add(n*2+1,i+n,inf,-b[i][0]-b[i][1]);
		add(n*2+2,i+n,inf,b[i][0]-b[i][1]);
		add(n*2+3,i+n,inf,-b[i][0]+b[i][1]);
		add(n*2+4,i+n,inf,b[i][0]+b[i][1]);
	}
	printf("%lld\n",SPFA::solve());
}
```