#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
const int N=1050;
const int M=2*N;
const ll inf=9e18;
struct Edge
{
	int u,v,c;ll w;
	Edge(){}
	Edge(int a, int b, int d, ll e):u(a),v(b),c(d),w(e){}
};
vector<Edge> edges;
vector<int> G[M];
void AddEdge(int u, int v, int c, ll w)
{
	G[u].pb(edges.size());
	edges.pb(Edge(u,v,c,w));
	G[v].pb(edges.size());
	edges.pb(Edge(v,u,0,-w));
}
int s,t;
ll dist[M];
int go[M];
bool in[M];
bool BFS()
{
	for(int i=s;i<=t;i++) dist[i]=-inf;
	dist[s]=0;
	queue<int> q;
	q.push(s);
	in[s]=1;
	while(q.size())
	{
		int u=q.front();
		q.pop();
		in[u]=0;
		//printf("u:%i dist[u]:%i\n",u,dist[u]);
		for(int e:G[u])
		{
			int v=edges[e].v;
			int c=edges[e].c;
			ll w=edges[e].w;
			//if(c>0) printf("Edge:%i %i %i %lld\n",u,v,c,w);
			if(c>0 && dist[v]<dist[u]+w)
			{
				dist[v]=dist[u]+w;
				go[v]=e;
				if(!in[v]) q.push(v),in[v]=1;
			}
		}
	}
	return dist[t]!=-inf;
}
ll MaxCostMaxFlow()
{
	ll ans=0;
	while(BFS())
	{
		int MaxCut=1e9+7;
		for(int i=t;i!=s;i=edges[go[i]].u) MaxCut=min(MaxCut,edges[go[i]].c);
		for(int i=t;i!=s;i=edges[go[i]].u) edges[go[i]].c-=MaxCut,edges[go[i]^1].c+=MaxCut;
		ans+=dist[t]*MaxCut;
	}
	return ans;
}
int main()
{
	int n,i,x,y,cnt;
	scanf("%i",&n);
	s=0;t=2*n+5;
	for(i=1;i<=n;i++)
	{
		scanf("%i %i %i",&x,&y,&cnt);
		AddEdge(s,i,cnt,0);
		AddEdge(i,2*n+1,cnt,x+y);
		AddEdge(i,2*n+2,cnt,x-y);
		AddEdge(i,2*n+3,cnt,-x+y);
		AddEdge(i,2*n+4,cnt,-x-y);
	}
	for(i=n+1;i<=2*n;i++)
	{
		scanf("%i %i %i",&x,&y,&cnt);
		AddEdge(i,t,cnt,0);
		AddEdge(2*n+1,i,cnt,-x-y);
		AddEdge(2*n+2,i,cnt,-x+y);
		AddEdge(2*n+3,i,cnt,x-y);
		AddEdge(2*n+4,i,cnt,x+y);
	}
	printf("%lld\n",MaxCostMaxFlow());
	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 <bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
const int N=1050;
const int M=2*N;
const ll inf=9e18;
struct Edge
{
	int u,v,c;ll w;
	Edge(){}
	Edge(int a, int b, int d, ll e):u(a),v(b),c(d),w(e){}
};
vector<Edge> edges;
vector<int> G[M];
void AddEdge(int u, int v, int c, ll w)
{
	G[u].pb(edges.size());
	edges.pb(Edge(u,v,c,w));
	G[v].pb(edges.size());
	edges.pb(Edge(v,u,0,-w));
}
int s,t;
ll dist[M];
int go[M];
bool in[M];
bool BFS()
{
	for(int i=s;i<=t;i++) dist[i]=-inf;
	dist[s]=0;
	queue<int> q;
	q.push(s);
	in[s]=1;
	while(q.size())
	{
		int u=q.front();
		q.pop();
		in[u]=0;
		//printf("u:%i dist[u]:%i\n",u,dist[u]);
		for(int e:G[u])
		{
			int v=edges[e].v;
			int c=edges[e].c;
			ll w=edges[e].w;
			//if(c>0) printf("Edge:%i %i %i %lld\n",u,v,c,w);
			if(c>0 && dist[v]<dist[u]+w)
			{
				dist[v]=dist[u]+w;
				go[v]=e;
				if(!in[v]) q.push(v),in[v]=1;
			}
		}
	}
	return dist[t]!=-inf;
}
ll MaxCostMaxFlow()
{
	ll ans=0;
	while(BFS())
	{
		int MaxCut=1e9+7;
		for(int i=t;i!=s;i=edges[go[i]].u) MaxCut=min(MaxCut,edges[go[i]].c);
		for(int i=t;i!=s;i=edges[go[i]].u) edges[go[i]].c-=MaxCut,edges[go[i]^1].c+=MaxCut;
		ans+=dist[t]*MaxCut;
	}
	return ans;
}
int main()
{
	int n,i,x,y,cnt;
	scanf("%i",&n);
	s=0;t=2*n+5;
	for(i=1;i<=n;i++)
	{
		scanf("%i %i %i",&x,&y,&cnt);
		AddEdge(s,i,cnt,0);
		AddEdge(i,2*n+1,cnt,x+y);
		AddEdge(i,2*n+2,cnt,x-y);
		AddEdge(i,2*n+3,cnt,-x+y);
		AddEdge(i,2*n+4,cnt,-x-y);
	}
	for(i=n+1;i<=2*n;i++)
	{
		scanf("%i %i %i",&x,&y,&cnt);
		AddEdge(i,t,cnt,0);
		AddEdge(2*n+1,i,cnt,-x-y);
		AddEdge(2*n+2,i,cnt,-x+y);
		AddEdge(2*n+3,i,cnt,x-y);
		AddEdge(2*n+4,i,cnt,x+y);
	}
	printf("%lld\n",MaxCostMaxFlow());
	return 0;
}
```