#include <set>
#include <queue>
#include <cmath>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(x) cerr<<#x<<'='<<x<<endl
#define set0(x) memset(x,0,sizeof(x))
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pii;
template<typename T> void read(T &x){
	x = 0;char ch = getchar();ll f = 1;
	while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
	while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
const int maxv = 5050;
ll m,n,s,t,u,v,ca,co;
ll mm,mf;
ll dist[maxv],inq[maxv],prevv[maxv],preve[maxv];
struct edge{ll to,ca,cost,rev;};
vector<edge> G[maxv];
void add_edge(ll u,ll v,ll ca,ll co){
    G[u].push_back((edge){v,ca,co,(int)G[v].size()});
    G[v].push_back((edge){u,0,-co,(int)G[u].size()-1});
}
ll inf = 1e16;
int main(){
  	s = 1;t = 2;
	int allo = 3;
  	int t1 =allo++,t2 = allo++,t3 = allo++,t4 = allo++;
	
	read(n);
	for(int i=0;i<n;i++){
		ll rx,ry,rc;
		read(rx);read(ry);read(rc);
		add_edge(s,allo,rc,0);
		add_edge(allo,t1,rc,1e10-(rx+ry));
		add_edge(allo,t2,rc,1e10-(-rx+ry));
		add_edge(allo,t3,rc,1e10-(rx-ry));
		add_edge(allo,t4,rc,1e10-(-rx-ry));
		allo+=1;
	}
	for(int i=0;i<n;i++){
		ll rx,ry,rc;
		read(rx);read(ry);read(rc);
		add_edge(t1,allo,rc,1e10+(rx+ry));
		add_edge(t2,allo,rc,1e10+(-rx+ry));
		add_edge(t3,allo,rc,1e10+(rx-ry));
		add_edge(t4,allo,rc,1e10+(-rx-ry));
		add_edge(allo,t,rc,0);
		allo+=1;
	}
    while(1){
        queue<int> que;
        for(int i=0;i<=allo;i++)dist[i] = inf;
		memset(inq,0,sizeof(inq));
        dist[s] = 0;
        que.push(s);
        inq[s] = 1;
        while(!que.empty()){
            int cur = que.front();
            que.pop();
            inq[cur] = 0;
            for(int i=0;i<G[cur].size();i++){
                edge cw = G[cur][i];
                if(cw.ca<=0)continue;
                if(dist[cw.to]>cw.cost+dist[cur]){
                    dist[cw.to] = cw.cost+dist[cur];
                    prevv[cw.to] = cur;
                    preve[cw.to] = i;
                    if(!inq[cw.to]){
                        que.push(cw.to);
                        inq[cw.to] = 1;
                    }
                }
            }
        }
        //cout<<dist[t]<<endl;
        if(dist[t]==inf)break;
        ll md = inf,di = 0;
        for(int p = t;p!=s;p = prevv[p]){
            md = min(md,G[prevv[p]][preve[p]].ca);
            di += G[prevv[p]][preve[p]].cost;
        }
        mm+=di*md;
        mf+=md;
        for(int p = t;p!=s;p = prevv[p]){
            G[prevv[p]][preve[p]].ca-=md;
            G[G[prevv[p]][preve[p]].to][G[prevv[p]][preve[p]].rev].ca+=md;
        }
    }
    cout<<mf*20000000000-mm<<endl;
    return 0;
}



### Prompt

Generate 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 <set>
#include <queue>
#include <cmath>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(x) cerr<<#x<<'='<<x<<endl
#define set0(x) memset(x,0,sizeof(x))
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pii;
template<typename T> void read(T &x){
	x = 0;char ch = getchar();ll f = 1;
	while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
	while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
const int maxv = 5050;
ll m,n,s,t,u,v,ca,co;
ll mm,mf;
ll dist[maxv],inq[maxv],prevv[maxv],preve[maxv];
struct edge{ll to,ca,cost,rev;};
vector<edge> G[maxv];
void add_edge(ll u,ll v,ll ca,ll co){
    G[u].push_back((edge){v,ca,co,(int)G[v].size()});
    G[v].push_back((edge){u,0,-co,(int)G[u].size()-1});
}
ll inf = 1e16;
int main(){
  	s = 1;t = 2;
	int allo = 3;
  	int t1 =allo++,t2 = allo++,t3 = allo++,t4 = allo++;
	
	read(n);
	for(int i=0;i<n;i++){
		ll rx,ry,rc;
		read(rx);read(ry);read(rc);
		add_edge(s,allo,rc,0);
		add_edge(allo,t1,rc,1e10-(rx+ry));
		add_edge(allo,t2,rc,1e10-(-rx+ry));
		add_edge(allo,t3,rc,1e10-(rx-ry));
		add_edge(allo,t4,rc,1e10-(-rx-ry));
		allo+=1;
	}
	for(int i=0;i<n;i++){
		ll rx,ry,rc;
		read(rx);read(ry);read(rc);
		add_edge(t1,allo,rc,1e10+(rx+ry));
		add_edge(t2,allo,rc,1e10+(-rx+ry));
		add_edge(t3,allo,rc,1e10+(rx-ry));
		add_edge(t4,allo,rc,1e10+(-rx-ry));
		add_edge(allo,t,rc,0);
		allo+=1;
	}
    while(1){
        queue<int> que;
        for(int i=0;i<=allo;i++)dist[i] = inf;
		memset(inq,0,sizeof(inq));
        dist[s] = 0;
        que.push(s);
        inq[s] = 1;
        while(!que.empty()){
            int cur = que.front();
            que.pop();
            inq[cur] = 0;
            for(int i=0;i<G[cur].size();i++){
                edge cw = G[cur][i];
                if(cw.ca<=0)continue;
                if(dist[cw.to]>cw.cost+dist[cur]){
                    dist[cw.to] = cw.cost+dist[cur];
                    prevv[cw.to] = cur;
                    preve[cw.to] = i;
                    if(!inq[cw.to]){
                        que.push(cw.to);
                        inq[cw.to] = 1;
                    }
                }
            }
        }
        //cout<<dist[t]<<endl;
        if(dist[t]==inf)break;
        ll md = inf,di = 0;
        for(int p = t;p!=s;p = prevv[p]){
            md = min(md,G[prevv[p]][preve[p]].ca);
            di += G[prevv[p]][preve[p]].cost;
        }
        mm+=di*md;
        mf+=md;
        for(int p = t;p!=s;p = prevv[p]){
            G[prevv[p]][preve[p]].ca-=md;
            G[G[prevv[p]][preve[p]].to][G[prevv[p]][preve[p]].rev].ca+=md;
        }
    }
    cout<<mf*20000000000-mm<<endl;
    return 0;
}



```