#include<cstdio>
#include<cstring>
#include<iostream>

#define nn 2110
#define ll long long
using namespace std;

ll as,d[nn];
struct Edge{int to,old,w,c;}e[nn*10];
int tot,n,fm[nn],q[nn*100],S,T,l[nn],v[nn];

inline void Lnk(int x,int y,int w,int c){
    e[++tot].to=y,e[tot].old=l[x],l[x]=tot,e[tot].w=w,e[tot].c=c;
    e[++tot].to=x,e[tot].old=l[y],l[y]=tot,e[tot].w=-w,e[tot].c=0;
}
inline int Spfa(){
    //memset(d,-1,8*(T+1));
    for (int i=1;i<=T;i++) d[i]=-1e18;
    memset(v,0,4*(T+1));
    memset(fm,0,4*(T+1));
    int h=1,t=1;
    q[1]=S,d[S]=0,v[S]=1;
    while (h<=t){
        int nw=q[h];
        for (int i=l[nw];i;i=e[i].old)if(d[e[i].to]<d[nw]+e[i].w&&e[i].c){
            fm[e[i].to]=i;
            d[e[i].to]=d[nw]+e[i].w;
            if (!v[e[i].to]) q[++t]=e[i].to,v[e[i].to]=1;
        }
        h++,v[nw]=0;
    }
    return(fm[T]);
}
int flow=0;
inline ll Get(){
    int a=110;ll rs=0;
    for (int i=T;i!=S;i=e[fm[i]^1].to) 
        a=min(a,e[fm[i]].c);
    for (int i=T;i!=S;i=e[fm[i]^1].to)
        rs+=a*1ll*e[fm[i]].w,e[fm[i]].c-=a,e[fm[i]^1].c+=a;
    flow+=a;
    return(rs);
}

int main()
{
    scanf("%d",&n);
    S=n*2+6,T=n*2+7,tot=1;
    for (int i=1;i<=n;i++){
        int x,y,z;scanf("%d%d%d",&x,&y,&z);
        Lnk(S,i,0,z);
        Lnk(i,n*2+1,x+y,z);
        Lnk(i,n*2+2,x-y,z);
        Lnk(i,n*2+3,-x+y,z);
        Lnk(i,n*2+4,-x-y,z);
    }
    for (int i=1;i<=n;i++){
        int x,y,z;scanf("%d%d%d",&x,&y,&z);
        Lnk(i+n,T,0,z);
        Lnk(n*2+1,i+n,-x-y,z);
        Lnk(n*2+2,i+n,-x+y,z);
        Lnk(n*2+3,i+n,x-y,z);
        Lnk(n*2+4,i+n,x+y,z);
    }

    while (Spfa())
        as+=Get();
    printf("%lld\n",as);
}
### 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<cstdio>
#include<cstring>
#include<iostream>

#define nn 2110
#define ll long long
using namespace std;

ll as,d[nn];
struct Edge{int to,old,w,c;}e[nn*10];
int tot,n,fm[nn],q[nn*100],S,T,l[nn],v[nn];

inline void Lnk(int x,int y,int w,int c){
    e[++tot].to=y,e[tot].old=l[x],l[x]=tot,e[tot].w=w,e[tot].c=c;
    e[++tot].to=x,e[tot].old=l[y],l[y]=tot,e[tot].w=-w,e[tot].c=0;
}
inline int Spfa(){
    //memset(d,-1,8*(T+1));
    for (int i=1;i<=T;i++) d[i]=-1e18;
    memset(v,0,4*(T+1));
    memset(fm,0,4*(T+1));
    int h=1,t=1;
    q[1]=S,d[S]=0,v[S]=1;
    while (h<=t){
        int nw=q[h];
        for (int i=l[nw];i;i=e[i].old)if(d[e[i].to]<d[nw]+e[i].w&&e[i].c){
            fm[e[i].to]=i;
            d[e[i].to]=d[nw]+e[i].w;
            if (!v[e[i].to]) q[++t]=e[i].to,v[e[i].to]=1;
        }
        h++,v[nw]=0;
    }
    return(fm[T]);
}
int flow=0;
inline ll Get(){
    int a=110;ll rs=0;
    for (int i=T;i!=S;i=e[fm[i]^1].to) 
        a=min(a,e[fm[i]].c);
    for (int i=T;i!=S;i=e[fm[i]^1].to)
        rs+=a*1ll*e[fm[i]].w,e[fm[i]].c-=a,e[fm[i]^1].c+=a;
    flow+=a;
    return(rs);
}

int main()
{
    scanf("%d",&n);
    S=n*2+6,T=n*2+7,tot=1;
    for (int i=1;i<=n;i++){
        int x,y,z;scanf("%d%d%d",&x,&y,&z);
        Lnk(S,i,0,z);
        Lnk(i,n*2+1,x+y,z);
        Lnk(i,n*2+2,x-y,z);
        Lnk(i,n*2+3,-x+y,z);
        Lnk(i,n*2+4,-x-y,z);
    }
    for (int i=1;i<=n;i++){
        int x,y,z;scanf("%d%d%d",&x,&y,&z);
        Lnk(i+n,T,0,z);
        Lnk(n*2+1,i+n,-x-y,z);
        Lnk(n*2+2,i+n,-x+y,z);
        Lnk(n*2+3,i+n,x-y,z);
        Lnk(n*2+4,i+n,x+y,z);
    }

    while (Spfa())
        as+=Get();
    printf("%lld\n",as);
}
```