#include <algorithm>
// #include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <queue>
#include <cmath>
#include <set>
#include <map>
#include <bitset>
using namespace std;
typedef long long LL;
typedef double db;
const int oo=2139063143;
const int N=12010;
const int P=1000000007;
const db eps=1e-7;
#define pritnf printf
//char buf[1<<22],*p1=buf,*p2=buf,obuf[1<<22],*O=obuf;
//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
template <typename T>
inline void sc (T &x)
{
    x=0; static int p; p=1; static char c; c=getchar();
    while (!isdigit(c)) { if (c=='-') p=-1; c=getchar(); }
    while ( isdigit(c)) { x=(x<<1)+(x<<3)+(c-48); c=getchar(); }
    x*=p;
}
template <typename T>
inline void print (T x)
{
    if (x< 0) putchar('-'),x=-x;
    if (x>=10) print(x/10);
    putchar(x%10+'0');
}
template <typename T>
inline void pr (T x) { print(x),putchar('\n'); }
template <typename T1,typename T2>
void chkmax (T1 &A,T2 B) { A=A>=B?A:B; }
template <typename T1,typename T2>
void chkmin (T1 &A,T2 B) { A=A<=B?A:B; }
int n;
struct EDGE { int v,w,c,nx; }lb[N<<3]; int tot=1,top[N];
void add (int u,int v,int w,int c)
{
	lb[++tot]=(EDGE){v,w, c,top[u]},top[u]=tot;
	lb[++tot]=(EDGE){u,0,-c,top[v]},top[v]=tot;
}
int s,t;
LL dis[N]; int flow[N],pre[N],las[N]; bool in[N];
bool spfa ()
{
	for (int i=1; i<=t; i++)
		dis[i]=-1e18,flow[i]=pre[i]=las[i]=0,in[i]=0;
	queue <int> q;
	flow[s]=oo; dis[s]=0,q.push (s),in[s]=1;
	while (q.size ())
	{
		int u=q.front (); q.pop (),in[u]=0;
		for (int kb=top[u]; kb; kb=lb[kb].nx)
		{
			int v=lb[kb].v,w=lb[kb].w,c=lb[kb].c;
			if (w&&dis[v]< dis[u]+c)
			{
				dis[v]=dis[u]+c;
				flow[v]=min (flow[u],w);
				pre[v]=u,las[v]=kb;
				if (!in[v]) q.push (v),in[v]=1;
			}
		}
	}
	return pre[t];
}
int maflow; LL macost;
void mcmf ()
{
	while (spfa ())
	{
		int u=t,fl=flow[t]; maflow+=fl,macost+=dis[t]*fl;
		while (u!=s) lb[las[u]].w-=fl,lb[las[u]^1].w+=fl,u=pre[u];
	}
}
struct LY { int x,y,w,id; }a[N],b[N];
const int dx[4]={1,-1,1,-1},dy[4]={1,1,-1,-1}; int id[4];
int main ()
{
    // freopen (".in","r",stdin);
    // freopen (".out","w",stdout);
	sc(n);
	for (int i=1; i<=n; i++)
		sc(a[i].x),sc(a[i].y),sc(a[i].w);
	for (int i=1; i<=n; i++)
		sc(b[i].x),sc(b[i].y),sc(b[i].w);
	s=n*2+4+1,t=s+1;
	for (int i=1; i<=n; i++)
		add (s,i,a[i].w,0),add (i+n,t,b[i].w,0);
	for (int i=1; i<=4; i++)
		id[i-1]=n*2+i;
	for (int i=1; i<=n; i++)
		for (int j=0; j<=3; j++)
			add (i,id[j],oo,dx[j]*a[i].x+dy[j]*a[i].y),add (id[j],i+n,oo,-(dx[j]*b[i].x+dy[j]*b[i].y));
	mcmf ();
	pr(macost);

    return 0;
}

### 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 <algorithm>
// #include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <queue>
#include <cmath>
#include <set>
#include <map>
#include <bitset>
using namespace std;
typedef long long LL;
typedef double db;
const int oo=2139063143;
const int N=12010;
const int P=1000000007;
const db eps=1e-7;
#define pritnf printf
//char buf[1<<22],*p1=buf,*p2=buf,obuf[1<<22],*O=obuf;
//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
template <typename T>
inline void sc (T &x)
{
    x=0; static int p; p=1; static char c; c=getchar();
    while (!isdigit(c)) { if (c=='-') p=-1; c=getchar(); }
    while ( isdigit(c)) { x=(x<<1)+(x<<3)+(c-48); c=getchar(); }
    x*=p;
}
template <typename T>
inline void print (T x)
{
    if (x< 0) putchar('-'),x=-x;
    if (x>=10) print(x/10);
    putchar(x%10+'0');
}
template <typename T>
inline void pr (T x) { print(x),putchar('\n'); }
template <typename T1,typename T2>
void chkmax (T1 &A,T2 B) { A=A>=B?A:B; }
template <typename T1,typename T2>
void chkmin (T1 &A,T2 B) { A=A<=B?A:B; }
int n;
struct EDGE { int v,w,c,nx; }lb[N<<3]; int tot=1,top[N];
void add (int u,int v,int w,int c)
{
	lb[++tot]=(EDGE){v,w, c,top[u]},top[u]=tot;
	lb[++tot]=(EDGE){u,0,-c,top[v]},top[v]=tot;
}
int s,t;
LL dis[N]; int flow[N],pre[N],las[N]; bool in[N];
bool spfa ()
{
	for (int i=1; i<=t; i++)
		dis[i]=-1e18,flow[i]=pre[i]=las[i]=0,in[i]=0;
	queue <int> q;
	flow[s]=oo; dis[s]=0,q.push (s),in[s]=1;
	while (q.size ())
	{
		int u=q.front (); q.pop (),in[u]=0;
		for (int kb=top[u]; kb; kb=lb[kb].nx)
		{
			int v=lb[kb].v,w=lb[kb].w,c=lb[kb].c;
			if (w&&dis[v]< dis[u]+c)
			{
				dis[v]=dis[u]+c;
				flow[v]=min (flow[u],w);
				pre[v]=u,las[v]=kb;
				if (!in[v]) q.push (v),in[v]=1;
			}
		}
	}
	return pre[t];
}
int maflow; LL macost;
void mcmf ()
{
	while (spfa ())
	{
		int u=t,fl=flow[t]; maflow+=fl,macost+=dis[t]*fl;
		while (u!=s) lb[las[u]].w-=fl,lb[las[u]^1].w+=fl,u=pre[u];
	}
}
struct LY { int x,y,w,id; }a[N],b[N];
const int dx[4]={1,-1,1,-1},dy[4]={1,1,-1,-1}; int id[4];
int main ()
{
    // freopen (".in","r",stdin);
    // freopen (".out","w",stdout);
	sc(n);
	for (int i=1; i<=n; i++)
		sc(a[i].x),sc(a[i].y),sc(a[i].w);
	for (int i=1; i<=n; i++)
		sc(b[i].x),sc(b[i].y),sc(b[i].w);
	s=n*2+4+1,t=s+1;
	for (int i=1; i<=n; i++)
		add (s,i,a[i].w,0),add (i+n,t,b[i].w,0);
	for (int i=1; i<=4; i++)
		id[i-1]=n*2+i;
	for (int i=1; i<=n; i++)
		for (int j=0; j<=3; j++)
			add (i,id[j],oo,dx[j]*a[i].x+dy[j]*a[i].y),add (id[j],i+n,oo,-(dx[j]*b[i].x+dy[j]*b[i].y));
	mcmf ();
	pr(macost);

    return 0;
}

```