#include<cstdio>
#include<cstring>
#define M 20002
#define N 2019
int A[N],B[N],C[N],S,T,a[N],b[M],c[M],d[M],e[M],g[N],i,l,n,q[N],r,t;long long f[N];bool h[N];
inline void add(int u,int v,int x,int y){b[++t]=a[u],c[a[u]=t]=v,d[t]=x,e[t]=y,b[++t]=a[v],c[a[v]=t]=u,e[t]=-y;}
int u,v,w;long long x;
inline bool bfs()
{
	memset(f,0x3f,T+5<<3),f[q[l=0]=S]=0,r=1;
	while(l!=r)for(h[u=q[l++]]=false,l=l==N?0:l,i=a[u];i;i=b[i])if(d[i]&&f[u]+e[i]<f[c[i]])
	{
		f[c[i]]=f[u]+e[i],g[c[i]]=i;
		if(!h[c[i]])h[q[r++]=c[i]]=true,r=r==N?0:r;
	}
	if(f[T]==0x3f3f3f3f3f3f3f3fll)return false;
	for(u=10,i=T;i!=S;i=c[g[i]^1])u=d[g[i]]<u?d[g[i]]:u;
	for(x-=u*f[i=T];i!=S;i=c[g[i]^1])d[g[i]]-=u,d[g[i]^1]+=u;
	return true;
}
int main()
{
	scanf("%d",&n),T=n<<1|1;
	for(i=t=1;i<=n;i++)scanf("%d%d%d",&u,&v,&w),add(S,i,w,0),add(i,T+1,10,u+v),add(i,T+2,10,u-v),add(i,T+3,10,v-u),add(i,T+4,10,-u-v);
	for(i=n+1;i<T;i++)scanf("%d%d%d",&u,&v,&w),add(T+1,i,10,-u-v),add(T+2,i,10,v-u),add(T+3,i,10,u-v),add(T+4,i,10,u+v),add(i,T,w,0);
	while(bfs());
	return 0*printf("%lld\n",x);
}
### Prompt

Your task is to create 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<cstdio>
#include<cstring>
#define M 20002
#define N 2019
int A[N],B[N],C[N],S,T,a[N],b[M],c[M],d[M],e[M],g[N],i,l,n,q[N],r,t;long long f[N];bool h[N];
inline void add(int u,int v,int x,int y){b[++t]=a[u],c[a[u]=t]=v,d[t]=x,e[t]=y,b[++t]=a[v],c[a[v]=t]=u,e[t]=-y;}
int u,v,w;long long x;
inline bool bfs()
{
	memset(f,0x3f,T+5<<3),f[q[l=0]=S]=0,r=1;
	while(l!=r)for(h[u=q[l++]]=false,l=l==N?0:l,i=a[u];i;i=b[i])if(d[i]&&f[u]+e[i]<f[c[i]])
	{
		f[c[i]]=f[u]+e[i],g[c[i]]=i;
		if(!h[c[i]])h[q[r++]=c[i]]=true,r=r==N?0:r;
	}
	if(f[T]==0x3f3f3f3f3f3f3f3fll)return false;
	for(u=10,i=T;i!=S;i=c[g[i]^1])u=d[g[i]]<u?d[g[i]]:u;
	for(x-=u*f[i=T];i!=S;i=c[g[i]^1])d[g[i]]-=u,d[g[i]^1]+=u;
	return true;
}
int main()
{
	scanf("%d",&n),T=n<<1|1;
	for(i=t=1;i<=n;i++)scanf("%d%d%d",&u,&v,&w),add(S,i,w,0),add(i,T+1,10,u+v),add(i,T+2,10,u-v),add(i,T+3,10,v-u),add(i,T+4,10,-u-v);
	for(i=n+1;i<T;i++)scanf("%d%d%d",&u,&v,&w),add(T+1,i,10,-u-v),add(T+2,i,10,v-u),add(T+3,i,10,u-v),add(T+4,i,10,u+v),add(i,T,w,0);
	while(bfs());
	return 0*printf("%lld\n",x);
}
```