#include<cstdio>
#include<cstring>
#define Z int
#define V void
#define L long long
#define S(n) scanf("%lld",&n)
#define P(n) printf("%lld",n)
#define M(f) memset(f,0x8f,sizeof f)
#define F(i,a,b) for(L i=a;i<=b;i++)
#define G(k,x) for(L k=las[x];k;k=nex[k])
#define W while
#define sm(a,b) ((a)<(b)?(a):(b))
#define I if
#define R return
#define nx 21000
#define inf 8102099357864587377
L n,x,y,b;
L las[nx],nex[nx],r[nx],v[nx],c[nx],l=1;
L p1,p2,p3,p4,s,t,stop,ans;
L d[nx+1],ex[nx],f[nx],to[nx],hd,tl,nw,fl;
V ins(L x,L y,L z,L p){r[++l]=y,nex[l]=las[x],v[las[x]=l]=z,c[l]=p;}
V link(L x,L y,L z,L p){ins(x,y,z,p),ins(y,x,0,-p);}
L spfa(){
	M(f),hd=0,f[d[tl=1]=s]=0;
	W(hd^tl){
		ex[d[hd=hd<nx?hd+1:1]]=0;
		G(k,d[hd])I(v[k])I(f[d[hd]]+c[k]>f[r[k]]){
			f[r[k]]=f[d[hd]]+c[to[r[k]]=k];
			I(!ex[r[k]])ex[d[tl=tl<nx?tl+1:1]=r[k]]=1;
		}
	}
	I(f[t]==-inf){stop=1;R 0;}
	nw=t,fl=inf;
	W(nw^s)fl=sm(fl,v[to[nw]]),nw=r[to[nw]^1];
	nw=t;
	W(nw^s)v[to[nw]]-=fl,v[to[nw]^1]+=fl,nw=r[to[nw]^1];
	R fl*f[t];
}
Z main(){
	S(n),p1=n+n+1,p2=p1+1,p3=p2+1,p4=p3+1,s=p4+1,t=s+1;
	F(i,1,n)S(x),S(y),S(b),link(s,i,b,0),
	link(i,p1,nx,-x-y),link(i,p2,nx,y-x),
	link(i,p3,nx,x-y),link(i,p4,nx,x+y);
	F(i,1,n)S(x),S(y),S(b),link(i+n,t,b,0),
	link(p1,i+n,nx,x+y),link(p2,i+n,nx,x-y),
	link(p3,i+n,nx,y-x),link(p4,i+n,nx,-x-y);
	W(!stop)ans+=spfa();
	P(ans);
}
### Prompt

Develop a solution in cpp to the problem described below:

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 Z int
#define V void
#define L long long
#define S(n) scanf("%lld",&n)
#define P(n) printf("%lld",n)
#define M(f) memset(f,0x8f,sizeof f)
#define F(i,a,b) for(L i=a;i<=b;i++)
#define G(k,x) for(L k=las[x];k;k=nex[k])
#define W while
#define sm(a,b) ((a)<(b)?(a):(b))
#define I if
#define R return
#define nx 21000
#define inf 8102099357864587377
L n,x,y,b;
L las[nx],nex[nx],r[nx],v[nx],c[nx],l=1;
L p1,p2,p3,p4,s,t,stop,ans;
L d[nx+1],ex[nx],f[nx],to[nx],hd,tl,nw,fl;
V ins(L x,L y,L z,L p){r[++l]=y,nex[l]=las[x],v[las[x]=l]=z,c[l]=p;}
V link(L x,L y,L z,L p){ins(x,y,z,p),ins(y,x,0,-p);}
L spfa(){
	M(f),hd=0,f[d[tl=1]=s]=0;
	W(hd^tl){
		ex[d[hd=hd<nx?hd+1:1]]=0;
		G(k,d[hd])I(v[k])I(f[d[hd]]+c[k]>f[r[k]]){
			f[r[k]]=f[d[hd]]+c[to[r[k]]=k];
			I(!ex[r[k]])ex[d[tl=tl<nx?tl+1:1]=r[k]]=1;
		}
	}
	I(f[t]==-inf){stop=1;R 0;}
	nw=t,fl=inf;
	W(nw^s)fl=sm(fl,v[to[nw]]),nw=r[to[nw]^1];
	nw=t;
	W(nw^s)v[to[nw]]-=fl,v[to[nw]^1]+=fl,nw=r[to[nw]^1];
	R fl*f[t];
}
Z main(){
	S(n),p1=n+n+1,p2=p1+1,p3=p2+1,p4=p3+1,s=p4+1,t=s+1;
	F(i,1,n)S(x),S(y),S(b),link(s,i,b,0),
	link(i,p1,nx,-x-y),link(i,p2,nx,y-x),
	link(i,p3,nx,x-y),link(i,p4,nx,x+y);
	F(i,1,n)S(x),S(y),S(b),link(i+n,t,b,0),
	link(p1,i+n,nx,x+y),link(p2,i+n,nx,x-y),
	link(p3,i+n,nx,y-x),link(p4,i+n,nx,-x-y);
	W(!stop)ans+=spfa();
	P(ans);
}
```