#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define pu putchar
#define pun putchar('\n')
int n,nn,u,v,w,e,ki,i,sx,qd,zh,q1,q2,q3,q4,x[30009],p[2000000],q[2009],z[2009],t[30009],li[30009],r[30009];
ll cx,g[50],hf[30009],d[2009];
bool b[2009],bb[2009];
int sri(){char c=getchar();int x=0;while(c<'0')c=getchar();while(c>='0')x=(x<<3)+(x<<1)+c-'0',c=getchar();return x;}
ll srl(){char c=getchar();ll x=0;while(c<'0')c=getchar();while(c>='0')x=(x<<3)+(x<<1)+c-'0',c=getchar();return x;}
void sc(ll x){int l=0;if(x==0)g[l=1]=0;while(x)g[++l]=x%10,x/=10;for(;l;l--)pu(g[l]+'0');}
void jia(int u,int v,int rr,ll h){
	x[++ki]=z[u];z[u]=ki,t[ki]=v;r[ki]=rr,li[ki]=0;hf[ki]=h;
	x[++ki]=z[v];z[v]=ki,t[ki]=u;r[ki]=0,li[ki]=0;hf[ki]=-h;
}
bool spfa(){
	for(int i=1;i<=zh;i++)bb[i]=0;
	u=0;w=1;p[w]=qd;b[qd]=1;d[qd]=0;bb[qd]=1;
	while(u!=w){
		u++;if(u&sx)u=1;v=p[u];b[v]=0;
		for(int i=z[v];i;i=x[i]){
			e=t[i];
			if(li[i]<r[i]&&(bb[e]==0||d[v]+hf[i]<d[e])){
				d[e]=d[v]+hf[i];q[e]=i;bb[e]=1;
				if(!b[e]){
					w++;if(w&sx)w=1;p[w]=e;b[e]=1;
				}
			}
		}
	}return bb[zh];
}
void ekspfa(){
	while(spfa()){int zz=n;
		for(int u=zh;u!=qd;u=t[q[u]^1])zz=min(zz,r[q[u]]-li[q[u]]);
		for(int u=zh;u!=qd;u=t[q[u]^1])li[q[u]]+=zz,li[q[u]^1]-=zz,cx+=zz*hf[q[u]];
	}
}
int main(){
//	freopen("1.in","r",stdin);
//	freopen("1.out","w",stdout);
	n=sri();nn=n<<1;q1=nn+1,q2=nn+2,q3=nn+3,q4=nn+4,qd=nn+5;zh=nn+6;ki=1;sx=1<<20;
	for(i=1;i<=n;i++){
		u=sri();v=sri();w=sri();jia(qd,i,w,0);
		jia(i,q1,w,-u-v);
		jia(i,q2,w,-u+v);
		jia(i,q3,w,u-v);
		jia(i,q4,w,u+v);
	}
	for(i=1;i<=n;i++){
		u=sri();v=sri();w=sri();jia(i+n,zh,w,0);
		jia(q1,i+n,w,u+v);
		jia(q2,i+n,w,u-v);
		jia(q3,i+n,w,-u+v);
		jia(q4,i+n,w,-u-v);
	}
    ekspfa();
    sc(-cx);
    return 0;
}
### 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>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define pu putchar
#define pun putchar('\n')
int n,nn,u,v,w,e,ki,i,sx,qd,zh,q1,q2,q3,q4,x[30009],p[2000000],q[2009],z[2009],t[30009],li[30009],r[30009];
ll cx,g[50],hf[30009],d[2009];
bool b[2009],bb[2009];
int sri(){char c=getchar();int x=0;while(c<'0')c=getchar();while(c>='0')x=(x<<3)+(x<<1)+c-'0',c=getchar();return x;}
ll srl(){char c=getchar();ll x=0;while(c<'0')c=getchar();while(c>='0')x=(x<<3)+(x<<1)+c-'0',c=getchar();return x;}
void sc(ll x){int l=0;if(x==0)g[l=1]=0;while(x)g[++l]=x%10,x/=10;for(;l;l--)pu(g[l]+'0');}
void jia(int u,int v,int rr,ll h){
	x[++ki]=z[u];z[u]=ki,t[ki]=v;r[ki]=rr,li[ki]=0;hf[ki]=h;
	x[++ki]=z[v];z[v]=ki,t[ki]=u;r[ki]=0,li[ki]=0;hf[ki]=-h;
}
bool spfa(){
	for(int i=1;i<=zh;i++)bb[i]=0;
	u=0;w=1;p[w]=qd;b[qd]=1;d[qd]=0;bb[qd]=1;
	while(u!=w){
		u++;if(u&sx)u=1;v=p[u];b[v]=0;
		for(int i=z[v];i;i=x[i]){
			e=t[i];
			if(li[i]<r[i]&&(bb[e]==0||d[v]+hf[i]<d[e])){
				d[e]=d[v]+hf[i];q[e]=i;bb[e]=1;
				if(!b[e]){
					w++;if(w&sx)w=1;p[w]=e;b[e]=1;
				}
			}
		}
	}return bb[zh];
}
void ekspfa(){
	while(spfa()){int zz=n;
		for(int u=zh;u!=qd;u=t[q[u]^1])zz=min(zz,r[q[u]]-li[q[u]]);
		for(int u=zh;u!=qd;u=t[q[u]^1])li[q[u]]+=zz,li[q[u]^1]-=zz,cx+=zz*hf[q[u]];
	}
}
int main(){
//	freopen("1.in","r",stdin);
//	freopen("1.out","w",stdout);
	n=sri();nn=n<<1;q1=nn+1,q2=nn+2,q3=nn+3,q4=nn+4,qd=nn+5;zh=nn+6;ki=1;sx=1<<20;
	for(i=1;i<=n;i++){
		u=sri();v=sri();w=sri();jia(qd,i,w,0);
		jia(i,q1,w,-u-v);
		jia(i,q2,w,-u+v);
		jia(i,q3,w,u-v);
		jia(i,q4,w,u+v);
	}
	for(i=1;i<=n;i++){
		u=sri();v=sri();w=sri();jia(i+n,zh,w,0);
		jia(q1,i+n,w,u+v);
		jia(q2,i+n,w,u-v);
		jia(q3,i+n,w,-u+v);
		jia(q4,i+n,w,-u-v);
	}
    ekspfa();
    sc(-cx);
    return 0;
}
```