#define George_Plover
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define MAXN 3000
#define MAXM 3000000
#define INF 2147483647
using namespace std;
int n,m,tx,ty,tz,T0,T1,T2,T3;
int S,T;
int pre[MAXN],tot=1,to[MAXM],lin[MAXM],cost[MAXM],flow[MAXM],in[MAXN];
LL dis[MAXN];
int bx[MAXN],by[MAXN];
bool vis[MAXN];
queue <int> q;
void add(int x,int y,int v,int f)
{
    tot++;
    lin[tot]=pre[x];
    pre[x]=tot;
    to[tot]=y;
    cost[tot]=v;
    flow[tot]=f;
    
    tot++;
    lin[tot]=pre[y];
    pre[y]=tot;
    to[tot]=x;
    cost[tot]=-v;
    flow[tot]=0;
}
bool SPFA()
{
    for(int i=0;i<MAXN;i++)
    {
        dis[i]=(1ll<<61);
    }
    memset(in,0,sizeof(in));
    dis[S]=0;
    q.push(S);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        vis[u]=0;
        
        for(int i=pre[u];i;i=lin[i])
        {
            int v=to[i];
            if(flow[i]&&dis[v]>dis[u]+cost[i])
            {
                dis[v]=dis[u]+cost[i];
                in[v]=i;
                if(!vis[v])
                {
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
    return dis[T]!=dis[0];
}

LL adjust()
{
    LL ret=2147483647;
    for(int i=T;i!=S;i=to[in[i]^1])
    {
        ret=(ret<flow[in[i]]?ret:flow[in[i]]);
    }
    for(int i=T;i!=S;i=to[in[i]^1])
    {
        flow[in[i]]-=ret;
        flow[in[i]^1]+=ret;
    }
    return ret*dis[T];
}

int main()
{
    scanf("%d",&n);
    S=n+n+2;
    T=S+1;T0=T+1;T1=T0+1;T2=T1+1;T3=T2+1;
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d%d",&tx,&ty,&tz);
        bx[i]=tx;
        by[i]=ty;
        add(S,i,0,tz);
        add(i,T0,tx+ty,INF);
        add(i,T1,-tx+ty,INF);
        add(i,T2,tx-ty,INF);
        add(i,T3,-tx-ty,INF);
    }
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d%d",&tx,&ty,&tz);
        add(i+n,T,0,tz);
        add(T0,n+i,-tx-ty,INF);
        add(T1,n+i,tx-ty,INF);
        add(T2,n+i,-tx+ty,INF);
        add(T3,n+i,tx+ty,INF);
    }
    
    LL ans=0;
    
    while(SPFA())
    {
        ans+=adjust();
    }
    
    cout<<-ans<<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
#define George_Plover
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define MAXN 3000
#define MAXM 3000000
#define INF 2147483647
using namespace std;
int n,m,tx,ty,tz,T0,T1,T2,T3;
int S,T;
int pre[MAXN],tot=1,to[MAXM],lin[MAXM],cost[MAXM],flow[MAXM],in[MAXN];
LL dis[MAXN];
int bx[MAXN],by[MAXN];
bool vis[MAXN];
queue <int> q;
void add(int x,int y,int v,int f)
{
    tot++;
    lin[tot]=pre[x];
    pre[x]=tot;
    to[tot]=y;
    cost[tot]=v;
    flow[tot]=f;
    
    tot++;
    lin[tot]=pre[y];
    pre[y]=tot;
    to[tot]=x;
    cost[tot]=-v;
    flow[tot]=0;
}
bool SPFA()
{
    for(int i=0;i<MAXN;i++)
    {
        dis[i]=(1ll<<61);
    }
    memset(in,0,sizeof(in));
    dis[S]=0;
    q.push(S);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        vis[u]=0;
        
        for(int i=pre[u];i;i=lin[i])
        {
            int v=to[i];
            if(flow[i]&&dis[v]>dis[u]+cost[i])
            {
                dis[v]=dis[u]+cost[i];
                in[v]=i;
                if(!vis[v])
                {
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
    return dis[T]!=dis[0];
}

LL adjust()
{
    LL ret=2147483647;
    for(int i=T;i!=S;i=to[in[i]^1])
    {
        ret=(ret<flow[in[i]]?ret:flow[in[i]]);
    }
    for(int i=T;i!=S;i=to[in[i]^1])
    {
        flow[in[i]]-=ret;
        flow[in[i]^1]+=ret;
    }
    return ret*dis[T];
}

int main()
{
    scanf("%d",&n);
    S=n+n+2;
    T=S+1;T0=T+1;T1=T0+1;T2=T1+1;T3=T2+1;
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d%d",&tx,&ty,&tz);
        bx[i]=tx;
        by[i]=ty;
        add(S,i,0,tz);
        add(i,T0,tx+ty,INF);
        add(i,T1,-tx+ty,INF);
        add(i,T2,tx-ty,INF);
        add(i,T3,-tx-ty,INF);
    }
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d%d",&tx,&ty,&tz);
        add(i+n,T,0,tz);
        add(T0,n+i,-tx-ty,INF);
        add(T1,n+i,tx-ty,INF);
        add(T2,n+i,-tx+ty,INF);
        add(T3,n+i,tx+ty,INF);
    }
    
    LL ans=0;
    
    while(SPFA())
    {
        ans+=adjust();
    }
    
    cout<<-ans<<endl;
    return 0;
}

```