#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
typedef long long LL;
const LL MAX=1e17;
const LL N=2100;
struct qq
{
    LL x,y,z,z1,last;
}e[N*10];LL num,last[N];
void init (LL x,LL y,LL z,LL z1)
{
    num++;
    e[num].x=x;e[num].y=y;e[num].z=z;e[num].z1=z1;
    e[num].last=last[x];
    last[x]=num;
    swap(x,y);z=0;z1=-z1;
    num++;
    e[num].x=x;e[num].y=y;e[num].z=z;e[num].z1=z1;
    e[num].last=last[x];
    last[x]=num;
}
LL from[N];
LL f[N];
bool in[N];
LL st,ed;
bool SPFA ()
{
    for (LL u=1;u<=ed;u++) from[u]=-1,f[u]=-MAX,in[u]=false;
    queue<int> q;q.push(st);
    f[st]=0;in[st]=true;
    while (!q.empty())
    {
        LL x=q.front();q.pop();
        for (LL u=last[x];u!=-1;u=e[u].last)
        {
            LL y=e[u].y;
            if (e[u].z>0&&f[y]<f[x]+e[u].z1)
            {
                f[y]=f[x]+e[u].z1;
                from[y]=u;
                if (in[y]==false)
                {
                    in[y]=true;
                    q.push(y);
                }
            }
        }
        in[x]=false;
    }
    return from[ed]!=-1;
}
LL ans=0;
void lalal ()
{
	//printf("YES!\n");
    LL x=from[ed],t=MAX;
    while (x!=-1)	{t=min(t,e[x].z);x=from[e[x].x];}
    x=from[ed];
    while (x!=-1)	{ans=ans+t*e[x].z1;e[x].z-=t;e[x^1].z+=t;x=from[e[x].x];}
    return ;
}
int main()
{
	num=1;memset(last,-1,sizeof(last));
	LL n; 
	scanf("%lld",&n);st=2*n+10+1;ed=st+1;
	LL cnt=0;
	for (LL u=1;u<=n;u++)
	{
		LL x,y,c;
		scanf("%lld%lld%lld",&x,&y,&c);
		init(st,u,c,0);
		init(u,2*n+1,MAX,x+y);
		init(u,2*n+2,MAX,x-y);
		init(u,2*n+3,MAX,-x+y);
		init(u,2*n+4,MAX,-x-y);
	}
	for (LL u=1;u<=n;u++)
	{
		LL x,y,c;
		scanf("%lld%lld%lld",&x,&y,&c);
		init(u+n,ed,c,0);
		init(2*n+1+4,u+n,MAX,x+y);
		init(2*n+2+4,u+n,MAX,x-y);
		init(2*n+3+4,u+n,MAX,-x+y);
		init(2*n+4+4,u+n,MAX,-x-y);
	}
	init(2*n+1,2*n+4+4,MAX,0);
	init(2*n+2,2*n+3+4,MAX,0);
	init(2*n+3,2*n+2+4,MAX,0);
	init(2*n+4,2*n+1+4,MAX,0);
	while (SPFA()) lalal();
	printf("%lld\n",ans);
	return 0;
}
### Prompt

In cpp, your task is to solve 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<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
typedef long long LL;
const LL MAX=1e17;
const LL N=2100;
struct qq
{
    LL x,y,z,z1,last;
}e[N*10];LL num,last[N];
void init (LL x,LL y,LL z,LL z1)
{
    num++;
    e[num].x=x;e[num].y=y;e[num].z=z;e[num].z1=z1;
    e[num].last=last[x];
    last[x]=num;
    swap(x,y);z=0;z1=-z1;
    num++;
    e[num].x=x;e[num].y=y;e[num].z=z;e[num].z1=z1;
    e[num].last=last[x];
    last[x]=num;
}
LL from[N];
LL f[N];
bool in[N];
LL st,ed;
bool SPFA ()
{
    for (LL u=1;u<=ed;u++) from[u]=-1,f[u]=-MAX,in[u]=false;
    queue<int> q;q.push(st);
    f[st]=0;in[st]=true;
    while (!q.empty())
    {
        LL x=q.front();q.pop();
        for (LL u=last[x];u!=-1;u=e[u].last)
        {
            LL y=e[u].y;
            if (e[u].z>0&&f[y]<f[x]+e[u].z1)
            {
                f[y]=f[x]+e[u].z1;
                from[y]=u;
                if (in[y]==false)
                {
                    in[y]=true;
                    q.push(y);
                }
            }
        }
        in[x]=false;
    }
    return from[ed]!=-1;
}
LL ans=0;
void lalal ()
{
	//printf("YES!\n");
    LL x=from[ed],t=MAX;
    while (x!=-1)	{t=min(t,e[x].z);x=from[e[x].x];}
    x=from[ed];
    while (x!=-1)	{ans=ans+t*e[x].z1;e[x].z-=t;e[x^1].z+=t;x=from[e[x].x];}
    return ;
}
int main()
{
	num=1;memset(last,-1,sizeof(last));
	LL n; 
	scanf("%lld",&n);st=2*n+10+1;ed=st+1;
	LL cnt=0;
	for (LL u=1;u<=n;u++)
	{
		LL x,y,c;
		scanf("%lld%lld%lld",&x,&y,&c);
		init(st,u,c,0);
		init(u,2*n+1,MAX,x+y);
		init(u,2*n+2,MAX,x-y);
		init(u,2*n+3,MAX,-x+y);
		init(u,2*n+4,MAX,-x-y);
	}
	for (LL u=1;u<=n;u++)
	{
		LL x,y,c;
		scanf("%lld%lld%lld",&x,&y,&c);
		init(u+n,ed,c,0);
		init(2*n+1+4,u+n,MAX,x+y);
		init(2*n+2+4,u+n,MAX,x-y);
		init(2*n+3+4,u+n,MAX,-x+y);
		init(2*n+4+4,u+n,MAX,-x-y);
	}
	init(2*n+1,2*n+4+4,MAX,0);
	init(2*n+2,2*n+3+4,MAX,0);
	init(2*n+3,2*n+2+4,MAX,0);
	init(2*n+4,2*n+1+4,MAX,0);
	while (SPFA()) lalal();
	printf("%lld\n",ans);
	return 0;
}
```