#include <queue>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

const int N=2.5e3+5;
const int M=2e4+5;
const int inf=1e9+7;

template <typename _Tp> inline void IN(_Tp&x) {
	char ch;bool flag=0;x=0;
	while(ch=getchar(),!isdigit(ch)) if(ch=='-') flag=1;
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	if(flag) x=-x;
}

namespace NetworkFlow {
	int S,T,cnt=1,vis[N],head[N];ll dis[N];
	struct Pre {int last,edge;} pre[N];
	struct Edge {int nxt,to,val;ll cot;} G[M];
	inline void addedge(int u,int v,int val,ll cot) {
		G[++cnt]=(Edge){head[u],v,val,cot},head[u]=cnt,
		G[++cnt]=(Edge){head[v],u,0,-cot},head[v]=cnt;
	}
	queue <int> q;
	inline bool Spfa() {
		memset(pre,0,sizeof(pre)),memset(vis,0,sizeof(vis));
		for(int i=1;i<=T;++i) dis[i]=-1e17;
		q.push(S),vis[S]=true,dis[S]=0;
		while(!q.empty()) {
			int u=q.front(),v;q.pop();
			for(int i=head[u];i;i=G[i].nxt)
				if(v=G[i].to,G[i].val>0&&dis[v]<dis[u]+G[i].cot) {
					dis[v]=dis[u]+G[i].cot,pre[v]=(Pre){u,i};
					if(!vis[v]) q.push(v),vis[v]=true;
				}
			vis[u]=false;
		}return dis[T]!=-1e17;
	}
	inline ll EK() {
		int tmp=0;ll cost=0;
		while(Spfa()) {
			tmp=inf;
			for(int i=T;i!=S;i=pre[i].last) tmp=min(tmp,G[pre[i].edge].val);
			for(int i=T;i!=S;i=pre[i].last) G[pre[i].edge].val-=tmp,G[pre[i].edge^1].val+=tmp;
			cost+=1ll*tmp*dis[T];
		}return cost;
	}
}using namespace NetworkFlow;

int n,P1,P2,P3,P4,rx[N],ry[N],rc[N],bx[N],by[N],bc[N];
int main() {
	IN(n),P1=2*n+1,P2=P1+1,P3=P2+1,P4=P3+1,S=P4+1,T=S+1;
	for(int i=1;i<=n;++i) IN(rx[i]),IN(ry[i]),IN(rc[i]);
	for(int i=1;i<=n;++i) IN(bx[i]),IN(by[i]),IN(bc[i]);
	for(int i=1;i<=n;++i) addedge(S,i,rc[i],0),addedge(i+n,T,bc[i],0);
	for(int i=1;i<=n;++i)
		addedge(i,P1,inf,rx[i]+ry[i]),addedge(P1,i+n,inf,-by[i]-bx[i]),
		addedge(i,P2,inf,rx[i]-ry[i]),addedge(P2,i+n,inf,by[i]-bx[i]),
		addedge(i,P3,inf,ry[i]-rx[i]),addedge(P3,i+n,inf,bx[i]-by[i]),
		addedge(i,P4,inf,-ry[i]-rx[i]),addedge(P4,i+n,inf,bx[i]+by[i]);
	printf("%lld\n",EK());
	return 0;
}
### Prompt

In Cpp, your task is to solve 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 <queue>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

const int N=2.5e3+5;
const int M=2e4+5;
const int inf=1e9+7;

template <typename _Tp> inline void IN(_Tp&x) {
	char ch;bool flag=0;x=0;
	while(ch=getchar(),!isdigit(ch)) if(ch=='-') flag=1;
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	if(flag) x=-x;
}

namespace NetworkFlow {
	int S,T,cnt=1,vis[N],head[N];ll dis[N];
	struct Pre {int last,edge;} pre[N];
	struct Edge {int nxt,to,val;ll cot;} G[M];
	inline void addedge(int u,int v,int val,ll cot) {
		G[++cnt]=(Edge){head[u],v,val,cot},head[u]=cnt,
		G[++cnt]=(Edge){head[v],u,0,-cot},head[v]=cnt;
	}
	queue <int> q;
	inline bool Spfa() {
		memset(pre,0,sizeof(pre)),memset(vis,0,sizeof(vis));
		for(int i=1;i<=T;++i) dis[i]=-1e17;
		q.push(S),vis[S]=true,dis[S]=0;
		while(!q.empty()) {
			int u=q.front(),v;q.pop();
			for(int i=head[u];i;i=G[i].nxt)
				if(v=G[i].to,G[i].val>0&&dis[v]<dis[u]+G[i].cot) {
					dis[v]=dis[u]+G[i].cot,pre[v]=(Pre){u,i};
					if(!vis[v]) q.push(v),vis[v]=true;
				}
			vis[u]=false;
		}return dis[T]!=-1e17;
	}
	inline ll EK() {
		int tmp=0;ll cost=0;
		while(Spfa()) {
			tmp=inf;
			for(int i=T;i!=S;i=pre[i].last) tmp=min(tmp,G[pre[i].edge].val);
			for(int i=T;i!=S;i=pre[i].last) G[pre[i].edge].val-=tmp,G[pre[i].edge^1].val+=tmp;
			cost+=1ll*tmp*dis[T];
		}return cost;
	}
}using namespace NetworkFlow;

int n,P1,P2,P3,P4,rx[N],ry[N],rc[N],bx[N],by[N],bc[N];
int main() {
	IN(n),P1=2*n+1,P2=P1+1,P3=P2+1,P4=P3+1,S=P4+1,T=S+1;
	for(int i=1;i<=n;++i) IN(rx[i]),IN(ry[i]),IN(rc[i]);
	for(int i=1;i<=n;++i) IN(bx[i]),IN(by[i]),IN(bc[i]);
	for(int i=1;i<=n;++i) addedge(S,i,rc[i],0),addedge(i+n,T,bc[i],0);
	for(int i=1;i<=n;++i)
		addedge(i,P1,inf,rx[i]+ry[i]),addedge(P1,i+n,inf,-by[i]-bx[i]),
		addedge(i,P2,inf,rx[i]-ry[i]),addedge(P2,i+n,inf,by[i]-bx[i]),
		addedge(i,P3,inf,ry[i]-rx[i]),addedge(P3,i+n,inf,bx[i]-by[i]),
		addedge(i,P4,inf,-ry[i]-rx[i]),addedge(P4,i+n,inf,bx[i]+by[i]);
	printf("%lld\n",EK());
	return 0;
}
```