#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2105,M=N*10,inf=1<<30;
int head[N],nxt[M],to[M],wei[M],cost[M],s,t,tot=1,n;
ll dis[N],ans; bool vis[N];
void addedge(int u, int v, int w, int c)
{
	nxt[++tot]=head[u],head[u]=tot,to[tot]=v,wei[tot]=w,cost[tot]=-c;
	nxt[++tot]=head[v],head[v]=tot,to[tot]=u,wei[tot]=0,cost[tot]=c;
}
bool spfa()
{
	memset(vis,0,sizeof(bool)*(t+3));
	memset(dis,0x3f,sizeof(ll)*(t+3));
	queue<int> q;
	dis[t]=0,q.push(t);
	while(!q.empty())
	{
		int u=q.front();
		vis[u]=false,q.pop();
		for(int e=head[u];e;e=nxt[e])
			if(wei[e^1]&&dis[to[e]]>dis[u]-cost[e])
			{
				dis[to[e]]=dis[u]-cost[e];
				if(!vis[to[e]]) vis[to[e]]=true,q.push(to[e]);
			}
	}
	return dis[s]!=dis[0];
}
int dfs(int u, int mx)
{
	vis[u]=true;
	if(u==t) return mx;
	int l=mx;
	for(int e=head[u];e&&l;e=nxt[e])
		if(!vis[to[e]]&&wei[e]>0&&dis[to[e]]==dis[u]-cost[e])
		{
			int f=dfs(to[e],min(l,wei[e]));
			ans+=1ll*f*cost[e];
			l-=f,wei[e]-=f,wei[e^1]+=f;
		}
	return mx-l;
}
ll exec()
{
	while(spfa())
		do
		{
			memset(vis,0,sizeof(bool)*(t+3));
			dfs(s,inf);
		} while(vis[t]);
	return -ans;
}
int main()
{
	scanf("%d",&n);
	s=2*n+5,t=s+1;
	for(int i=1,x,y,c;i<=n;++i)
	{
		scanf("%d%d%d",&x,&y,&c);
		addedge(s,i,c,0);
		addedge(i,2*n+1,inf,x+y);
		addedge(i,2*n+2,inf,x-y);
		addedge(i,2*n+3,inf,-x+y);
		addedge(i,2*n+4,inf,-x-y);
	}
	for(int i=1,x,y,c;i<=n;++i)
	{
		scanf("%d%d%d",&x,&y,&c);
		addedge(2*n+1,i+n,inf,-x-y);
		addedge(2*n+2,i+n,inf,-x+y);
		addedge(2*n+3,i+n,inf,x-y);
		addedge(2*n+4,i+n,inf,x+y);
		addedge(i+n,t,c,0);
	}
	printf("%lld\n",exec());
}

### Prompt

Please create a solution in Cpp 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;
typedef long long ll;
const int N=2105,M=N*10,inf=1<<30;
int head[N],nxt[M],to[M],wei[M],cost[M],s,t,tot=1,n;
ll dis[N],ans; bool vis[N];
void addedge(int u, int v, int w, int c)
{
	nxt[++tot]=head[u],head[u]=tot,to[tot]=v,wei[tot]=w,cost[tot]=-c;
	nxt[++tot]=head[v],head[v]=tot,to[tot]=u,wei[tot]=0,cost[tot]=c;
}
bool spfa()
{
	memset(vis,0,sizeof(bool)*(t+3));
	memset(dis,0x3f,sizeof(ll)*(t+3));
	queue<int> q;
	dis[t]=0,q.push(t);
	while(!q.empty())
	{
		int u=q.front();
		vis[u]=false,q.pop();
		for(int e=head[u];e;e=nxt[e])
			if(wei[e^1]&&dis[to[e]]>dis[u]-cost[e])
			{
				dis[to[e]]=dis[u]-cost[e];
				if(!vis[to[e]]) vis[to[e]]=true,q.push(to[e]);
			}
	}
	return dis[s]!=dis[0];
}
int dfs(int u, int mx)
{
	vis[u]=true;
	if(u==t) return mx;
	int l=mx;
	for(int e=head[u];e&&l;e=nxt[e])
		if(!vis[to[e]]&&wei[e]>0&&dis[to[e]]==dis[u]-cost[e])
		{
			int f=dfs(to[e],min(l,wei[e]));
			ans+=1ll*f*cost[e];
			l-=f,wei[e]-=f,wei[e^1]+=f;
		}
	return mx-l;
}
ll exec()
{
	while(spfa())
		do
		{
			memset(vis,0,sizeof(bool)*(t+3));
			dfs(s,inf);
		} while(vis[t]);
	return -ans;
}
int main()
{
	scanf("%d",&n);
	s=2*n+5,t=s+1;
	for(int i=1,x,y,c;i<=n;++i)
	{
		scanf("%d%d%d",&x,&y,&c);
		addedge(s,i,c,0);
		addedge(i,2*n+1,inf,x+y);
		addedge(i,2*n+2,inf,x-y);
		addedge(i,2*n+3,inf,-x+y);
		addedge(i,2*n+4,inf,-x-y);
	}
	for(int i=1,x,y,c;i<=n;++i)
	{
		scanf("%d%d%d",&x,&y,&c);
		addedge(2*n+1,i+n,inf,-x-y);
		addedge(2*n+2,i+n,inf,-x+y);
		addedge(2*n+3,i+n,inf,x-y);
		addedge(2*n+4,i+n,inf,x+y);
		addedge(i+n,t,c,0);
	}
	printf("%lld\n",exec());
}

```