#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 2010
#define S 0
#define T 2001
#define red(x) (x)
#define blue(x) (n+(x))
#define id(x) (2002+(x))
#define int long long
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,p[N],d[N],q[N],pre[N],t=-1,ans;
bool flag[N];
struct data{int x,y,c,v[4];
}a[N],b[N];
struct data2{int to,nxt,cap,flow,cost;
}edge[N<<5];
void addedge(int x,int y,int z,int cost)
{
    t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].cap=z,edge[t].flow=0,edge[t].cost=cost,p[x]=t;
    t++;edge[t].to=x,edge[t].nxt=p[y],edge[t].cap=0,edge[t].flow=0,edge[t].cost=-cost,p[y]=t;
}
int inc(int &x){x++;if (x>2007) x-=2007;return x;}
bool spfa()
{
    memset(d,60,sizeof(d));d[S]=0;
    memset(flag,0,sizeof(flag));
    int head=0,tail=1;q[1]=S;
    do
    {
        int x=q[inc(head)];flag[x]=0;
        for (int i=p[x];~i;i=edge[i].nxt)
        if (edge[i].flow<edge[i].cap&&d[x]+edge[i].cost<d[edge[i].to])
        {
            d[edge[i].to]=d[x]+edge[i].cost;
            pre[edge[i].to]=i;
            if (!flag[edge[i].to]) q[inc(tail)]=edge[i].to,flag[edge[i].to]=1;
        }
    }while (head!=tail);
    return d[T]<10000000000000000ll;
}
void ekspfa()
{
    while (spfa())
    {
        int v=n;
        for (int i=T;i!=S;i=edge[pre[i]^1].to)
        v=min(v,edge[pre[i]].cap-edge[pre[i]].flow);
        for (int i=T;i!=S;i=edge[pre[i]^1].to)
        edge[pre[i]].flow+=v,edge[pre[i]^1].flow-=v,ans+=v*edge[pre[i]].cost;
    }
}
signed main()
{
    n=read();memset(p,255,sizeof(p));
    for (int i=1;i<=n;i++)
    {
        a[i].x=read(),a[i].y=read(),a[i].c=read();
        a[i].v[0]=a[i].x+a[i].y;
        a[i].v[1]=a[i].x-a[i].y;
        a[i].v[2]=-a[i].x+a[i].y;
        a[i].v[3]=-a[i].x-a[i].y;
    }
    for (int i=1;i<=n;i++)
    {
        b[i].x=read(),b[i].y=read(),b[i].c=read();
        b[i].v[0]=-b[i].x-b[i].y;
        b[i].v[1]=-b[i].x+b[i].y;
        b[i].v[2]=b[i].x-b[i].y;
        b[i].v[3]=b[i].x+b[i].y;
    }
    for (int i=1;i<=n;i++)
    {
        addedge(S,red(i),a[i].c,0),addedge(blue(i),T,b[i].c,0);
        for (int j=0;j<4;j++)
        addedge(red(i),id(j),a[i].c,-a[i].v[j]),addedge(id(j),blue(i),b[i].c,-b[i].v[j]);
    }
    ekspfa();
    cout<<-ans;
    return 0;
    //NOTICE LONG LONG!!!!!
}
### Prompt

Please formulate 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 inf 1000000010
#define N 2010
#define S 0
#define T 2001
#define red(x) (x)
#define blue(x) (n+(x))
#define id(x) (2002+(x))
#define int long long
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,p[N],d[N],q[N],pre[N],t=-1,ans;
bool flag[N];
struct data{int x,y,c,v[4];
}a[N],b[N];
struct data2{int to,nxt,cap,flow,cost;
}edge[N<<5];
void addedge(int x,int y,int z,int cost)
{
    t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].cap=z,edge[t].flow=0,edge[t].cost=cost,p[x]=t;
    t++;edge[t].to=x,edge[t].nxt=p[y],edge[t].cap=0,edge[t].flow=0,edge[t].cost=-cost,p[y]=t;
}
int inc(int &x){x++;if (x>2007) x-=2007;return x;}
bool spfa()
{
    memset(d,60,sizeof(d));d[S]=0;
    memset(flag,0,sizeof(flag));
    int head=0,tail=1;q[1]=S;
    do
    {
        int x=q[inc(head)];flag[x]=0;
        for (int i=p[x];~i;i=edge[i].nxt)
        if (edge[i].flow<edge[i].cap&&d[x]+edge[i].cost<d[edge[i].to])
        {
            d[edge[i].to]=d[x]+edge[i].cost;
            pre[edge[i].to]=i;
            if (!flag[edge[i].to]) q[inc(tail)]=edge[i].to,flag[edge[i].to]=1;
        }
    }while (head!=tail);
    return d[T]<10000000000000000ll;
}
void ekspfa()
{
    while (spfa())
    {
        int v=n;
        for (int i=T;i!=S;i=edge[pre[i]^1].to)
        v=min(v,edge[pre[i]].cap-edge[pre[i]].flow);
        for (int i=T;i!=S;i=edge[pre[i]^1].to)
        edge[pre[i]].flow+=v,edge[pre[i]^1].flow-=v,ans+=v*edge[pre[i]].cost;
    }
}
signed main()
{
    n=read();memset(p,255,sizeof(p));
    for (int i=1;i<=n;i++)
    {
        a[i].x=read(),a[i].y=read(),a[i].c=read();
        a[i].v[0]=a[i].x+a[i].y;
        a[i].v[1]=a[i].x-a[i].y;
        a[i].v[2]=-a[i].x+a[i].y;
        a[i].v[3]=-a[i].x-a[i].y;
    }
    for (int i=1;i<=n;i++)
    {
        b[i].x=read(),b[i].y=read(),b[i].c=read();
        b[i].v[0]=-b[i].x-b[i].y;
        b[i].v[1]=-b[i].x+b[i].y;
        b[i].v[2]=b[i].x-b[i].y;
        b[i].v[3]=b[i].x+b[i].y;
    }
    for (int i=1;i<=n;i++)
    {
        addedge(S,red(i),a[i].c,0),addedge(blue(i),T,b[i].c,0);
        for (int j=0;j<4;j++)
        addedge(red(i),id(j),a[i].c,-a[i].v[j]),addedge(id(j),blue(i),b[i].c,-b[i].v[j]);
    }
    ekspfa();
    cout<<-ans;
    return 0;
    //NOTICE LONG LONG!!!!!
}
```