#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define ll long long
using namespace std;
int rdn()
{
  int ret=0;bool fx=1;char ch=getchar();
  while(ch>'9'||ch<'0'){if(ch=='-')fx=0;ch=getchar();}
  while(ch>='0'&&ch<='9')ret=ret*10+ch-'0',ch=getchar();
  return fx?ret:-ret;
}
int Mn(int a,int b){return a<b?a:b;}
const int N=2010,M=2e4+5; const ll INF=(ll)(2e9+5)*M;
int n,S,T,tot,hd[N],xnt=1,to[M],nxt[M],cap[M],w[M]; ll ans;
int info[N],pre[N];ll dis[N]; bool ins[N];
queue<int> q;
void add(int x,int y,int c,int z)
{
  to[++xnt]=y;nxt[xnt]=hd[x];hd[x]=xnt;cap[xnt]=c;w[xnt]=z;
  to[++xnt]=x;nxt[xnt]=hd[y];hd[y]=xnt;cap[xnt]=0;w[xnt]=-z;
}
bool spfa()
{
  for(int i=1;i<=tot;i++)dis[i]=-INF;
  dis[S]=0; info[S]=N*10; info[T]=0; q.push(S);
  while(q.size())
    {
      int k=q.front(); q.pop(); ins[k]=0;
      for(int i=hd[k],v;i;i=nxt[i])
	if(cap[i]&&dis[v=to[i]]<dis[k]+w[i])
	  {
	    dis[v]=dis[k]+w[i];
	    pre[v]=i; info[v]=Mn(cap[i],info[k]);
	    if(!ins[v])ins[v]=1,q.push(v);
	  }
    }
  return info[T];
}
void ek()
{
  int tp=info[T];
  for(int i=pre[T];i;i=pre[to[i^1]])
    {
      cap[i]-=tp; cap[i^1]+=tp;
      ans+=(ll)w[i]*tp;
    }
}
int main()
{
  n=rdn(); tot=4; S=++tot; T=++tot;
  for(int i=1,x,y,c;i<=n;i++)
    {
      x=rdn();y=rdn();c=rdn();
      tot++; add(S,tot,c,0);
      add(tot,1,c,x+y); add(tot,2,c,x-y);
      add(tot,3,c,-x+y); add(tot,4,c,-x-y);
    }
  for(int i=1,x,y,c;i<=n;i++)
    {
      x=rdn();y=rdn();c=rdn();
      tot++; add(tot,T,c,0);
      add(1,tot,c,-x-y); add(2,tot,c,-x+y);
      add(3,tot,c,x-y); add(4,tot,c,x+y);
    }
  while(spfa())ek();
  printf("%lld\n",ans);
  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<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define ll long long
using namespace std;
int rdn()
{
  int ret=0;bool fx=1;char ch=getchar();
  while(ch>'9'||ch<'0'){if(ch=='-')fx=0;ch=getchar();}
  while(ch>='0'&&ch<='9')ret=ret*10+ch-'0',ch=getchar();
  return fx?ret:-ret;
}
int Mn(int a,int b){return a<b?a:b;}
const int N=2010,M=2e4+5; const ll INF=(ll)(2e9+5)*M;
int n,S,T,tot,hd[N],xnt=1,to[M],nxt[M],cap[M],w[M]; ll ans;
int info[N],pre[N];ll dis[N]; bool ins[N];
queue<int> q;
void add(int x,int y,int c,int z)
{
  to[++xnt]=y;nxt[xnt]=hd[x];hd[x]=xnt;cap[xnt]=c;w[xnt]=z;
  to[++xnt]=x;nxt[xnt]=hd[y];hd[y]=xnt;cap[xnt]=0;w[xnt]=-z;
}
bool spfa()
{
  for(int i=1;i<=tot;i++)dis[i]=-INF;
  dis[S]=0; info[S]=N*10; info[T]=0; q.push(S);
  while(q.size())
    {
      int k=q.front(); q.pop(); ins[k]=0;
      for(int i=hd[k],v;i;i=nxt[i])
	if(cap[i]&&dis[v=to[i]]<dis[k]+w[i])
	  {
	    dis[v]=dis[k]+w[i];
	    pre[v]=i; info[v]=Mn(cap[i],info[k]);
	    if(!ins[v])ins[v]=1,q.push(v);
	  }
    }
  return info[T];
}
void ek()
{
  int tp=info[T];
  for(int i=pre[T];i;i=pre[to[i^1]])
    {
      cap[i]-=tp; cap[i^1]+=tp;
      ans+=(ll)w[i]*tp;
    }
}
int main()
{
  n=rdn(); tot=4; S=++tot; T=++tot;
  for(int i=1,x,y,c;i<=n;i++)
    {
      x=rdn();y=rdn();c=rdn();
      tot++; add(S,tot,c,0);
      add(tot,1,c,x+y); add(tot,2,c,x-y);
      add(tot,3,c,-x+y); add(tot,4,c,-x-y);
    }
  for(int i=1,x,y,c;i<=n;i++)
    {
      x=rdn();y=rdn();c=rdn();
      tot++; add(tot,T,c,0);
      add(1,tot,c,-x-y); add(2,tot,c,-x+y);
      add(3,tot,c,x-y); add(4,tot,c,x+y);
    }
  while(spfa())ek();
  printf("%lld\n",ans);
  return 0;
}

```