#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int N=2333;
struct edge{int to,nxt,w,cost;}a[N<<5];
int n,S,T,hd[N],cnt=1,vis[N],pe[N];long long dis[N],ans;
queue<int>Q;
void link(int u,int v,int w,int cost){
	a[++cnt]=(edge){v,hd[u],w,cost};hd[u]=cnt;
	a[++cnt]=(edge){u,hd[v],0,-cost};hd[v]=cnt;
}
bool spfa(){
	memset(dis,-63,sizeof(dis));
	dis[S]=0;Q.push(S);
	while(!Q.empty()){
		int u=Q.front();Q.pop();
		for(int i=hd[u];i;i=a[i].nxt){
			int v=a[i].to;
			if(a[i].w&&dis[v]<dis[u]+a[i].cost){
				dis[v]=dis[u]+a[i].cost;pe[v]=i;
				if(!vis[v])vis[v]=1,Q.push(v);
			}
		}
		vis[u]=0;
	}
	if(dis[T]==dis[0])return false;
	int sum=1<<30;
	for(int i=T;i!=S;i=a[pe[i]^1].to)sum=min(sum,a[pe[i]].w);
	ans+=dis[T]*sum;
	for(int i=T;i!=S;i=a[pe[i]^1].to)a[pe[i]].w-=sum,a[pe[i]^1].w+=sum;
	return true;
}
int main(){
	scanf("%d",&n);S=n+n+1;T=S+1;
	for(int i=1,x,y,z;i<=n;++i){
		scanf("%d%d%d",&x,&y,&z);
		link(S,i,z,0);
		link(i,n+n+3,z,x+y);
		link(i,n+n+4,z,x-y);
		link(i,n+n+5,z,-x+y);
		link(i,n+n+6,z,-x-y);
	}
	for(int i=1,x,y,z;i<=n;++i){
		scanf("%d%d%d",&x,&y,&z);
		link(i+n,T,z,0);
		link(n+n+3,i+n,z,-x-y);
		link(n+n+4,i+n,z,-x+y);
		link(n+n+5,i+n,z,x-y);
		link(n+n+6,i+n,z,x+y);
	}
	while(spfa());
	printf("%lld\n",ans);
	return 0;
}

### Prompt

Your challenge is to write 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<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int N=2333;
struct edge{int to,nxt,w,cost;}a[N<<5];
int n,S,T,hd[N],cnt=1,vis[N],pe[N];long long dis[N],ans;
queue<int>Q;
void link(int u,int v,int w,int cost){
	a[++cnt]=(edge){v,hd[u],w,cost};hd[u]=cnt;
	a[++cnt]=(edge){u,hd[v],0,-cost};hd[v]=cnt;
}
bool spfa(){
	memset(dis,-63,sizeof(dis));
	dis[S]=0;Q.push(S);
	while(!Q.empty()){
		int u=Q.front();Q.pop();
		for(int i=hd[u];i;i=a[i].nxt){
			int v=a[i].to;
			if(a[i].w&&dis[v]<dis[u]+a[i].cost){
				dis[v]=dis[u]+a[i].cost;pe[v]=i;
				if(!vis[v])vis[v]=1,Q.push(v);
			}
		}
		vis[u]=0;
	}
	if(dis[T]==dis[0])return false;
	int sum=1<<30;
	for(int i=T;i!=S;i=a[pe[i]^1].to)sum=min(sum,a[pe[i]].w);
	ans+=dis[T]*sum;
	for(int i=T;i!=S;i=a[pe[i]^1].to)a[pe[i]].w-=sum,a[pe[i]^1].w+=sum;
	return true;
}
int main(){
	scanf("%d",&n);S=n+n+1;T=S+1;
	for(int i=1,x,y,z;i<=n;++i){
		scanf("%d%d%d",&x,&y,&z);
		link(S,i,z,0);
		link(i,n+n+3,z,x+y);
		link(i,n+n+4,z,x-y);
		link(i,n+n+5,z,-x+y);
		link(i,n+n+6,z,-x-y);
	}
	for(int i=1,x,y,z;i<=n;++i){
		scanf("%d%d%d",&x,&y,&z);
		link(i+n,T,z,0);
		link(n+n+3,i+n,z,-x-y);
		link(n+n+4,i+n,z,-x+y);
		link(n+n+5,i+n,z,x-y);
		link(n+n+6,i+n,z,x+y);
	}
	while(spfa());
	printf("%lld\n",ans);
	return 0;
}

```