#include<cstdio>
#include<queue>
#include<iostream>
#define RI register int
#define CI const int&
using namespace std;
typedef long long LL;
const int N=2100,M=20*N,INF=1e9;
int n,c,x,y,s,t;
namespace NF //Network_Flow
{
	struct edge
	{
		int to,nxt,v,c;
	}e[M]; queue <int> q; bool vis[N];
	int head[N],cnt=1,cap[N],pre[N],lst[N]; LL dis[N];
	#define to e[i].to
	inline bool SPFA(CI s,CI t)
	{
		RI i; for (i=s;i<=t;++i) dis[i]=-1e18,pre[i]=-1;
		q.push(s); dis[s]=0; cap[s]=INF; vis[s]=1; while (!q.empty())
		{
			int now=q.front(); q.pop(); vis[now]=0;
			for (i=head[now];i;i=e[i].nxt) if (e[i].v&&dis[now]+e[i].c>dis[to])
			{
				dis[to]=dis[now]+e[i].c; cap[to]=min(cap[now],e[i].v);
				pre[to]=now; lst[to]=i; if (!vis[to]) vis[to]=1,q.push(to);
			}
		}
		return ~pre[t];
		#undef to
	}
	inline void addedge(CI x,CI y,CI v,CI c)
	{
		e[++cnt]=(edge){y,head[x],v,c}; head[x]=cnt;
		e[++cnt]=(edge){x,head[y],0,-c}; head[y]=cnt;
	}
	inline LL MCMF(CI s,CI t)
	{
		LL ret=0; while (SPFA(s,t))
		{
			ret+=dis[t]*cap[t]; for (int nw=t;nw!=s;nw=pre[nw]) 
			e[lst[nw]].v-=cap[t],e[lst[nw]^1].v+=cap[t];
		}
		return ret;
	}
};
int main()
{
	RI i; for (scanf("%d",&n),s=0,t=2*n+5,i=1;i<=n;++i)
	scanf("%d%d%d",&x,&y,&c),NF::addedge(s,i,c,0),
	NF::addedge(i,2*n+1,INF,x+y),NF::addedge(i,2*n+2,INF,x-y),
	NF::addedge(i,2*n+3,INF,-x+y),NF::addedge(i,2*n+4,INF,-x-y);
	for (i=1;i<=n;++i)
	scanf("%d%d%d",&x,&y,&c),NF::addedge(n+i,t,c,0),
	NF::addedge(2*n+1,n+i,INF,-x-y),NF::addedge(2*n+2,n+i,INF,-x+y),
	NF::addedge(2*n+3,n+i,INF,x-y),NF::addedge(2*n+4,n+i,INF,x+y);
	return printf("%lld",NF::MCMF(s,t)),0;
}
### Prompt

Create a solution in Cpp for 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<queue>
#include<iostream>
#define RI register int
#define CI const int&
using namespace std;
typedef long long LL;
const int N=2100,M=20*N,INF=1e9;
int n,c,x,y,s,t;
namespace NF //Network_Flow
{
	struct edge
	{
		int to,nxt,v,c;
	}e[M]; queue <int> q; bool vis[N];
	int head[N],cnt=1,cap[N],pre[N],lst[N]; LL dis[N];
	#define to e[i].to
	inline bool SPFA(CI s,CI t)
	{
		RI i; for (i=s;i<=t;++i) dis[i]=-1e18,pre[i]=-1;
		q.push(s); dis[s]=0; cap[s]=INF; vis[s]=1; while (!q.empty())
		{
			int now=q.front(); q.pop(); vis[now]=0;
			for (i=head[now];i;i=e[i].nxt) if (e[i].v&&dis[now]+e[i].c>dis[to])
			{
				dis[to]=dis[now]+e[i].c; cap[to]=min(cap[now],e[i].v);
				pre[to]=now; lst[to]=i; if (!vis[to]) vis[to]=1,q.push(to);
			}
		}
		return ~pre[t];
		#undef to
	}
	inline void addedge(CI x,CI y,CI v,CI c)
	{
		e[++cnt]=(edge){y,head[x],v,c}; head[x]=cnt;
		e[++cnt]=(edge){x,head[y],0,-c}; head[y]=cnt;
	}
	inline LL MCMF(CI s,CI t)
	{
		LL ret=0; while (SPFA(s,t))
		{
			ret+=dis[t]*cap[t]; for (int nw=t;nw!=s;nw=pre[nw]) 
			e[lst[nw]].v-=cap[t],e[lst[nw]^1].v+=cap[t];
		}
		return ret;
	}
};
int main()
{
	RI i; for (scanf("%d",&n),s=0,t=2*n+5,i=1;i<=n;++i)
	scanf("%d%d%d",&x,&y,&c),NF::addedge(s,i,c,0),
	NF::addedge(i,2*n+1,INF,x+y),NF::addedge(i,2*n+2,INF,x-y),
	NF::addedge(i,2*n+3,INF,-x+y),NF::addedge(i,2*n+4,INF,-x-y);
	for (i=1;i<=n;++i)
	scanf("%d%d%d",&x,&y,&c),NF::addedge(n+i,t,c,0),
	NF::addedge(2*n+1,n+i,INF,-x-y),NF::addedge(2*n+2,n+i,INF,-x+y),
	NF::addedge(2*n+3,n+i,INF,x-y),NF::addedge(2*n+4,n+i,INF,x+y);
	return printf("%lld",NF::MCMF(s,t)),0;
}
```