#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define maxn 2010
#define maxm 100005
#define maxd 5000000
#define inf 2000000000ll
#define ll long long 
using namespace std;

int n,i,j,k,S,T;
int em,e[maxm],nx[maxm],ls[maxn],ec[maxm];
ll cost[maxm];

void insert(int x,int y,int z,ll c){
	em++;e[em]=y;nx[em]=ls[x];ls[x]=em;ec[em]=z;cost[em]=c;
	em++;e[em]=x;nx[em]=ls[y];ls[y]=em;ec[em]=0;cost[em]=-c;
}

ll ans,sum,dis[maxn];
int d[maxd],vis[maxn],fa[maxn];
int SPFA(){
	for(int i=1;i<=T;i++) dis[i]=-1e15,vis[i]=fa[i]=0;
	int t=0,w=1;d[1]=S,dis[S]=0,vis[S]=1;
	while (t<w){
		int x=d[++t]; vis[x]=0;
		for(int i=ls[x];i;i=nx[i]) if (ec[i]&&dis[x]+cost[i]>dis[e[i]]){
			dis[e[i]]=dis[x]+cost[i],fa[e[i]]=i;
			if (!vis[e[i]]) {
				vis[e[i]]=1,d[++w]=e[i];
				if (dis[d[w]]<dis[d[t+1]]) swap(d[w],d[t+1]);
			}
		}
	}
	return dis[T]>-1e15;
}

void maxflow(){
	while (SPFA()){
		int mi=inf;
		for(int x=T;x!=S;x=e[fa[x]^1]) mi=min(mi,ec[fa[x]]);
		sum+=mi;
		for(int x=T;x!=S;x=e[fa[x]^1]) 
			ec[fa[x]]-=mi,ec[fa[x]^1]+=mi,ans+=cost[fa[x]]*mi;
	}
}

int main(){
	scanf("%d",&n);
	S=2*n+5,T=2*n+6,em=1;
	for(i=1;i<=n;i++){
		int x,y,c,nm=2*n;scanf("%d%d%d",&x,&y,&c);
		insert(S,i,c,0);
		for(j=0;j<=1;j++) for(k=0;k<=1;k++)
			insert(i,++nm,inf,inf+x*(j?1:-1)+y*(k?1:-1));
	}
	for(i=1;i<=n;i++){
		int x,y,c,nm=2*n;scanf("%d%d%d",&x,&y,&c);
		insert(n+i,T,c,0);
		for(j=0;j<=1;j++) for(k=0;k<=1;k++)
			insert(++nm,n+i,inf,inf+x*(j?-1:1)+y*(k?-1:1));
	}
	maxflow();
	printf("%lld",ans-inf*sum*2);
}
### Prompt

Develop a solution in Cpp to the problem described below:

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<cmath>
#include<cstring>
#include<algorithm>
#define maxn 2010
#define maxm 100005
#define maxd 5000000
#define inf 2000000000ll
#define ll long long 
using namespace std;

int n,i,j,k,S,T;
int em,e[maxm],nx[maxm],ls[maxn],ec[maxm];
ll cost[maxm];

void insert(int x,int y,int z,ll c){
	em++;e[em]=y;nx[em]=ls[x];ls[x]=em;ec[em]=z;cost[em]=c;
	em++;e[em]=x;nx[em]=ls[y];ls[y]=em;ec[em]=0;cost[em]=-c;
}

ll ans,sum,dis[maxn];
int d[maxd],vis[maxn],fa[maxn];
int SPFA(){
	for(int i=1;i<=T;i++) dis[i]=-1e15,vis[i]=fa[i]=0;
	int t=0,w=1;d[1]=S,dis[S]=0,vis[S]=1;
	while (t<w){
		int x=d[++t]; vis[x]=0;
		for(int i=ls[x];i;i=nx[i]) if (ec[i]&&dis[x]+cost[i]>dis[e[i]]){
			dis[e[i]]=dis[x]+cost[i],fa[e[i]]=i;
			if (!vis[e[i]]) {
				vis[e[i]]=1,d[++w]=e[i];
				if (dis[d[w]]<dis[d[t+1]]) swap(d[w],d[t+1]);
			}
		}
	}
	return dis[T]>-1e15;
}

void maxflow(){
	while (SPFA()){
		int mi=inf;
		for(int x=T;x!=S;x=e[fa[x]^1]) mi=min(mi,ec[fa[x]]);
		sum+=mi;
		for(int x=T;x!=S;x=e[fa[x]^1]) 
			ec[fa[x]]-=mi,ec[fa[x]^1]+=mi,ans+=cost[fa[x]]*mi;
	}
}

int main(){
	scanf("%d",&n);
	S=2*n+5,T=2*n+6,em=1;
	for(i=1;i<=n;i++){
		int x,y,c,nm=2*n;scanf("%d%d%d",&x,&y,&c);
		insert(S,i,c,0);
		for(j=0;j<=1;j++) for(k=0;k<=1;k++)
			insert(i,++nm,inf,inf+x*(j?1:-1)+y*(k?1:-1));
	}
	for(i=1;i<=n;i++){
		int x,y,c,nm=2*n;scanf("%d%d%d",&x,&y,&c);
		insert(n+i,T,c,0);
		for(j=0;j<=1;j++) for(k=0;k<=1;k++)
			insert(++nm,n+i,inf,inf+x*(j?-1:1)+y*(k?-1:1));
	}
	maxflow();
	printf("%lld",ans-inf*sum*2);
}
```