#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1000005
#define inf 1e18
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

inline ll read()
{
	ll x=0,w=1; char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
	while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
	return w==1?x:-x;
}

struct node{ll from,to,cap,flow,cost;};
vector <node> edges;
vector <int> mp[maxn];
ll s,t,n,m,d[maxn],flow,cost,p[maxn],a[maxn];
bool inq[maxn];
queue <int> que;

inline void A(int u,int v,ll w,ll c)
{
	edges.push_back((node){u,v,w,0,c});
	edges.push_back((node){v,u,0,0,-c});
	int pp=edges.size();
	mp[u].push_back(pp-2);
	mp[v].push_back(pp-1);
}

inline bool spfa()
{
	for(int i=0;i<=t;i++) d[i]=inf,inq[i]=0,a[i]=inf;
	que.push(s); d[s]=0;
	while(!que.empty())
	{
		int u=que.front(); que.pop(); inq[u]=0;
		for(int i=0;i<mp[u].size();i++)
		{
			node &v=edges[mp[u][i]];
			if(v.cap-v.flow>0&&d[v.to]>d[u]+v.cost)
			{
				d[v.to]=d[u]+v.cost;
				a[v.to]=min(a[u],v.cap-v.flow);
				p[v.to]=mp[u][i];
				if(inq[v.to]==0) que.push(v.to),inq[v.to]=1;
			}
		}
	}
	if(d[t]==inf) return false;
	flow+=a[t];
	cost+=a[t]*d[t];
	for(int i=t;i!=s;i=edges[p[i]].from)
	{
		edges[p[i]].flow+=a[t];
		edges[p[i]^1].flow-=a[t];
	}
	return true;
}

int main()
{
	n=read(); s=0; t=2*n+5;
	rep(i,1,n)
	{
		ll a=read(),b=read(),c=read();
		A(s,i,c,0); A(i,2*n+1,inf,a+b); A(i,2*n+2,inf,a-b);
		A(i,2*n+3,inf,-a+b); A(i,2*n+4,inf,-a-b);
	}
	rep(i,1,n)
	{
		ll a=read(),b=read(),c=read();
		A(n+i,t,c,0); A(2*n+1,n+i,inf,-a-b); A(2*n+2,n+i,inf,-a+b);
		A(2*n+3,n+i,inf,a-b); A(2*n+4,n+i,inf,a+b);
	}
	while(spfa()); cout<<-cost<<endl;
	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
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1000005
#define inf 1e18
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

inline ll read()
{
	ll x=0,w=1; char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
	while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
	return w==1?x:-x;
}

struct node{ll from,to,cap,flow,cost;};
vector <node> edges;
vector <int> mp[maxn];
ll s,t,n,m,d[maxn],flow,cost,p[maxn],a[maxn];
bool inq[maxn];
queue <int> que;

inline void A(int u,int v,ll w,ll c)
{
	edges.push_back((node){u,v,w,0,c});
	edges.push_back((node){v,u,0,0,-c});
	int pp=edges.size();
	mp[u].push_back(pp-2);
	mp[v].push_back(pp-1);
}

inline bool spfa()
{
	for(int i=0;i<=t;i++) d[i]=inf,inq[i]=0,a[i]=inf;
	que.push(s); d[s]=0;
	while(!que.empty())
	{
		int u=que.front(); que.pop(); inq[u]=0;
		for(int i=0;i<mp[u].size();i++)
		{
			node &v=edges[mp[u][i]];
			if(v.cap-v.flow>0&&d[v.to]>d[u]+v.cost)
			{
				d[v.to]=d[u]+v.cost;
				a[v.to]=min(a[u],v.cap-v.flow);
				p[v.to]=mp[u][i];
				if(inq[v.to]==0) que.push(v.to),inq[v.to]=1;
			}
		}
	}
	if(d[t]==inf) return false;
	flow+=a[t];
	cost+=a[t]*d[t];
	for(int i=t;i!=s;i=edges[p[i]].from)
	{
		edges[p[i]].flow+=a[t];
		edges[p[i]^1].flow-=a[t];
	}
	return true;
}

int main()
{
	n=read(); s=0; t=2*n+5;
	rep(i,1,n)
	{
		ll a=read(),b=read(),c=read();
		A(s,i,c,0); A(i,2*n+1,inf,a+b); A(i,2*n+2,inf,a-b);
		A(i,2*n+3,inf,-a+b); A(i,2*n+4,inf,-a-b);
	}
	rep(i,1,n)
	{
		ll a=read(),b=read(),c=read();
		A(n+i,t,c,0); A(2*n+1,n+i,inf,-a-b); A(2*n+2,n+i,inf,-a+b);
		A(2*n+3,n+i,inf,a-b); A(2*n+4,n+i,inf,a+b);
	}
	while(spfa()); cout<<-cost<<endl;
	return 0;
}
```