#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
typedef long long LL;
const int N=6e3,inf=1e9;
struct edge{
    int to,nxt,cap,cost;
}e[N*10];
int n,cnt=1,head[N],T,iter[N];
LL dist[N],ans;
bool vis[N];
queue<int>q;
inline void addedge(int u,int v,int cost,int cap){
    e[++cnt]=(edge){v,head[u],cap,cost},head[u]=cnt;
    e[++cnt]=(edge){u,head[v],0,-cost},head[v]=cnt;
}
int dfs(int u,int f){
    if(u==T||!f)return f;
    vis[u]=1;
    int ret=0;
    for(int&i=iter[u];i;i=e[i].nxt)
    if(e[i].cap&&!vis[e[i].to]&&dist[e[i].to]==dist[u]+e[i].cost){
        int d=dfs(e[i].to,min(f,e[i].cap));
        e[i].cap-=d,e[i^1].cap+=d,f-=d,ret+=d;
        if(!f)break;
    }
    vis[u]=0;
    return ret;
}
bool MCMF(){
    for(int i=1;i<=T;++i)dist[i]=-1e18;
    for(q.push(0);!q.empty();){
        int u=q.front();
        q.pop();
        vis[u]=0;
        for(int i=head[u];i;i=e[i].nxt)
        if(e[i].cap&&dist[e[i].to]<dist[u]+e[i].cost){
            dist[e[i].to]=dist[u]+e[i].cost;
            if(!vis[e[i].to])
            vis[e[i].to]=1,q.push(e[i].to);
        }
    }
    if(dist[T]<-1e17)return 0;
    memcpy(iter,head,sizeof(int)*(T+2));
    while(int f=dfs(0,inf))ans+=dist[T]*f;
    return 1;
}
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin>>n;
    T=2*n+9;
    for(int i=1;i<=n;++i){
        int x,y,c;
        cin>>x>>y>>c;
        addedge(0,i,0,c);
        addedge(i,2*n+1,x+y,c);
        addedge(i,2*n+2,x-y,c);
        addedge(i,2*n+3,y-x,c);
        addedge(i,2*n+4,-x-y,c);
    }
    for(int i=n+1;i<=2*n;++i){
        int x,y,c;
        cin>>x>>y>>c;
        addedge(i,T,0,c);
        addedge(2*n+5,i,x+y,c);
        addedge(2*n+6,i,x-y,c);
        addedge(2*n+7,i,y-x,c);
        addedge(2*n+8,i,-x-y,c);
    }
    addedge(2*n+1,2*n+8,0,inf);
    addedge(2*n+2,2*n+7,0,inf);
    addedge(2*n+3,2*n+6,0,inf);
    addedge(2*n+4,2*n+5,0,inf);
    while(MCMF());
    cout<<ans<<'\n';
    return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
typedef long long LL;
const int N=6e3,inf=1e9;
struct edge{
    int to,nxt,cap,cost;
}e[N*10];
int n,cnt=1,head[N],T,iter[N];
LL dist[N],ans;
bool vis[N];
queue<int>q;
inline void addedge(int u,int v,int cost,int cap){
    e[++cnt]=(edge){v,head[u],cap,cost},head[u]=cnt;
    e[++cnt]=(edge){u,head[v],0,-cost},head[v]=cnt;
}
int dfs(int u,int f){
    if(u==T||!f)return f;
    vis[u]=1;
    int ret=0;
    for(int&i=iter[u];i;i=e[i].nxt)
    if(e[i].cap&&!vis[e[i].to]&&dist[e[i].to]==dist[u]+e[i].cost){
        int d=dfs(e[i].to,min(f,e[i].cap));
        e[i].cap-=d,e[i^1].cap+=d,f-=d,ret+=d;
        if(!f)break;
    }
    vis[u]=0;
    return ret;
}
bool MCMF(){
    for(int i=1;i<=T;++i)dist[i]=-1e18;
    for(q.push(0);!q.empty();){
        int u=q.front();
        q.pop();
        vis[u]=0;
        for(int i=head[u];i;i=e[i].nxt)
        if(e[i].cap&&dist[e[i].to]<dist[u]+e[i].cost){
            dist[e[i].to]=dist[u]+e[i].cost;
            if(!vis[e[i].to])
            vis[e[i].to]=1,q.push(e[i].to);
        }
    }
    if(dist[T]<-1e17)return 0;
    memcpy(iter,head,sizeof(int)*(T+2));
    while(int f=dfs(0,inf))ans+=dist[T]*f;
    return 1;
}
int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin>>n;
    T=2*n+9;
    for(int i=1;i<=n;++i){
        int x,y,c;
        cin>>x>>y>>c;
        addedge(0,i,0,c);
        addedge(i,2*n+1,x+y,c);
        addedge(i,2*n+2,x-y,c);
        addedge(i,2*n+3,y-x,c);
        addedge(i,2*n+4,-x-y,c);
    }
    for(int i=n+1;i<=2*n;++i){
        int x,y,c;
        cin>>x>>y>>c;
        addedge(i,T,0,c);
        addedge(2*n+5,i,x+y,c);
        addedge(2*n+6,i,x-y,c);
        addedge(2*n+7,i,y-x,c);
        addedge(2*n+8,i,-x-y,c);
    }
    addedge(2*n+1,2*n+8,0,inf);
    addedge(2*n+2,2*n+7,0,inf);
    addedge(2*n+3,2*n+6,0,inf);
    addedge(2*n+4,2*n+5,0,inf);
    while(MCMF());
    cout<<ans<<'\n';
    return 0;
}
```