#include<bits/stdc++.h>
using namespace std;

#define int long long 

void read(int &x) {
    x=0;int f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-f;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';x*=f;
}

void print(int x) {
    if(x<0) putchar('-'),x=-x;
    if(!x) return ;print(x/10),putchar(x%10+48);
}
void write(int x) {if(!x) putchar('0');else print(x);putchar('\n');}

#define lf double
#define ll long long 

#define pii pair<int,int >
#define vec vector<int >

#define pb push_back
#define mp make_pair
#define fr first
#define sc second

#define data asd09123jdf02i3h

#define FOR(i,l,r) for(int i=l,i##_r=r;i<=i##_r;i++)

const int maxn = 1e6+10;
const int inf = 1e12;
const lf eps = 1e-8;
const int mod = 1e9+7;

int head[maxn],tot=1,n,s,t,max_cost;

struct edge{int to,nxt,w,c;}e[maxn<<1];

void add(int u,int v,int w,int c) {e[++tot]=(edge){v,head[u],w,c},head[u]=tot;}
void ins(int u,int v,int w,int c) {add(u,v,w,c),add(v,u,0,-c);}

int dis[2040],pre[maxn],id[maxn],vis[maxn];

int gao() {
    memset(dis,-63,sizeof dis);
    dis[s]=0;queue<int > q;q.push(s);
    while(!q.empty()) {
        int x=q.front();q.pop();vis[x]=0;
        for(int i=head[x],v;i;i=e[i].nxt)
            if(e[i].w&&dis[v=e[i].to]<dis[x]+e[i].c) {
                dis[v]=dis[x]+e[i].c;
                pre[v]=x,id[v]=i;
                if(!vis[v]) vis[v]=1,q.push(v);
            }
    }

    if(dis[t]==dis[0]) return 0;
    int mn=inf,x=t;
    while(x!=s) mn=min(e[id[x]].w,mn),x=pre[x];
    x=t;
    while(x!=s) {
        max_cost+=e[id[x]].c*mn;
        e[id[x]].w-=mn,e[id[x]^1].w+=mn;
        x=pre[x];
    }
    return 1;
}

signed main() {
    read(n);s=n*2+1,t=s+1;
    int p1=t+1,p2=t+2,p3=t+3,p4=t+4;
    for(int i=1;i<=n;i++) {
        int c,x,y;read(x),read(y),read(c);
        ins(s,i,c,0);
        ins(i,p1,c,x+y);
        ins(i,p2,c,x-y);
        ins(i,p3,c,-x+y);
        ins(i,p4,c,-x-y);
    }
    for(int i=1;i<=n;i++) {
        int c,x,y;read(x),read(y),read(c);
        ins(i+n,t,c,0);
        ins(p1,i+n,c,-x-y);
        ins(p2,i+n,c,-x+y);
        ins(p3,i+n,c,x-y);
        ins(p4,i+n,c,x+y);
    }
    while(gao());
    write(max_cost);
    return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 int long long 

void read(int &x) {
    x=0;int f=1;char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-f;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';x*=f;
}

void print(int x) {
    if(x<0) putchar('-'),x=-x;
    if(!x) return ;print(x/10),putchar(x%10+48);
}
void write(int x) {if(!x) putchar('0');else print(x);putchar('\n');}

#define lf double
#define ll long long 

#define pii pair<int,int >
#define vec vector<int >

#define pb push_back
#define mp make_pair
#define fr first
#define sc second

#define data asd09123jdf02i3h

#define FOR(i,l,r) for(int i=l,i##_r=r;i<=i##_r;i++)

const int maxn = 1e6+10;
const int inf = 1e12;
const lf eps = 1e-8;
const int mod = 1e9+7;

int head[maxn],tot=1,n,s,t,max_cost;

struct edge{int to,nxt,w,c;}e[maxn<<1];

void add(int u,int v,int w,int c) {e[++tot]=(edge){v,head[u],w,c},head[u]=tot;}
void ins(int u,int v,int w,int c) {add(u,v,w,c),add(v,u,0,-c);}

int dis[2040],pre[maxn],id[maxn],vis[maxn];

int gao() {
    memset(dis,-63,sizeof dis);
    dis[s]=0;queue<int > q;q.push(s);
    while(!q.empty()) {
        int x=q.front();q.pop();vis[x]=0;
        for(int i=head[x],v;i;i=e[i].nxt)
            if(e[i].w&&dis[v=e[i].to]<dis[x]+e[i].c) {
                dis[v]=dis[x]+e[i].c;
                pre[v]=x,id[v]=i;
                if(!vis[v]) vis[v]=1,q.push(v);
            }
    }

    if(dis[t]==dis[0]) return 0;
    int mn=inf,x=t;
    while(x!=s) mn=min(e[id[x]].w,mn),x=pre[x];
    x=t;
    while(x!=s) {
        max_cost+=e[id[x]].c*mn;
        e[id[x]].w-=mn,e[id[x]^1].w+=mn;
        x=pre[x];
    }
    return 1;
}

signed main() {
    read(n);s=n*2+1,t=s+1;
    int p1=t+1,p2=t+2,p3=t+3,p4=t+4;
    for(int i=1;i<=n;i++) {
        int c,x,y;read(x),read(y),read(c);
        ins(s,i,c,0);
        ins(i,p1,c,x+y);
        ins(i,p2,c,x-y);
        ins(i,p3,c,-x+y);
        ins(i,p4,c,-x-y);
    }
    for(int i=1;i<=n;i++) {
        int c,x,y;read(x),read(y),read(c);
        ins(i+n,t,c,0);
        ins(p1,i+n,c,-x-y);
        ins(p2,i+n,c,-x+y);
        ins(p3,i+n,c,x-y);
        ins(p4,i+n,c,x+y);
    }
    while(gao());
    write(max_cost);
    return 0;
}

```