#include<bits/stdc++.h>
using namespace std;
long long n,m,s,t,x,y,z,g,top,head,tail,top1,top2,top3,min1,i,j,ans,ans2,
		  o=1000000000,zz[2100000+10],zz2[2100000+10],zz3[2100000+10],
		  to[2100000+10],nex[2100000+10],fir[2100000+10],fei[2100000+10],
		  liu[2100000+10],d[40000000+10],book[2100000+10],vis[2100000+10];
long long rx[1005],ry[1005],rc[1005],bx[1005],by[1005],bc[1005],a,b,c,e;
int lj(long long u,long long v,long long w,long long f){
	top++;
	to[top]=v;
	liu[top]=w;
	fei[top]=f;
	nex[top]=fir[u];
	fir[u]=top;
	return 0;
}
int lj2(long long u,long long v,long long w,long long f){
	lj(u,v,w,f);
	lj(v,u,0,-f);
	return 0;
}
int main(){
	scanf("%lld",&n);
	s=(n<<1)+1;
	t=s+1;
	a=t+1;
	b=a+1;
	c=b+1;
	e=c+1;
	for(i=1;i<=n;i++)
		scanf("%lld%lld%lld",&rx[i],&ry[i],&rc[i]);
	for(i=1;i<=n;i++)
		scanf("%lld%lld%lld",&bx[i],&by[i],&bc[i]);
	for(i=1;i<=n;i++){
		lj2(i,a,rc[i],rx[i]+ry[i]);
		lj2(i,b,rc[i],rx[i]-ry[i]);
		lj2(i,c,rc[i],-rx[i]+ry[i]);
		lj2(i,e,rc[i],-rx[i]-ry[i]);
	}
	for(i=1;i<=n;i++){
		lj2(a,i+n,bc[i],-bx[i]-by[i]);
		lj2(b,i+n,bc[i],-bx[i]+by[i]);
		lj2(c,i+n,bc[i],bx[i]-by[i]);
		lj2(e,i+n,bc[i],bx[i]+by[i]);
	}
	for(i=1;i<=n;i++)
		lj2(s,i,rc[i],0);
	for(i=1;i<=n;i++)
		lj2(i+n,t,bc[i],0);
	n=e;
	o*=o;
	while(1)
	{
		head=1;tail=1;
		d[1]=s;
		
		for(i=1;i<=n;i++)
			vis[i]=-o;
		vis[s]=0;
		book[s]=1;
		while(head<=tail){
			top2=d[head];
			top1=fir[top2];
			while(top1!=0){
				if(vis[top2]+fei[top1]>vis[to[top1]]&&liu[top1]!=0){
					vis[to[top1]]=vis[top2]+fei[top1];
					zz[to[top1]]=top2;
					zz2[to[top1]]=liu[top1];
					zz3[to[top1]]=top1;
					if(book[to[top1]]==0)
					{
						tail++;
						d[tail]=to[top1];
						book[to[top1]]=1;
					}
				}
				top1=nex[top1];
			}
			head++;
			book[top2]=0;
		}
		if(vis[t]<0)
			break;
		top1=t;
		min1=o;
		while(top1!=s){
			min1=min(min1,zz2[top1]);
			top1=zz[top1];
		}
		top1=t;
		ans+=min1;
		ans2+=min1*vis[t];
		while(top1!=s){
			top2=zz3[top1];
			if(top2%2==1)
				top3=top2+1;
			else
				top3=top2-1;
			top1=zz[top1];
			liu[top2]-=min1;
			liu[top3]+=min1;
		}
	}
	cout<<ans2<<endl;
	return 0;
}
### Prompt

Your challenge is to write 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<bits/stdc++.h>
using namespace std;
long long n,m,s,t,x,y,z,g,top,head,tail,top1,top2,top3,min1,i,j,ans,ans2,
		  o=1000000000,zz[2100000+10],zz2[2100000+10],zz3[2100000+10],
		  to[2100000+10],nex[2100000+10],fir[2100000+10],fei[2100000+10],
		  liu[2100000+10],d[40000000+10],book[2100000+10],vis[2100000+10];
long long rx[1005],ry[1005],rc[1005],bx[1005],by[1005],bc[1005],a,b,c,e;
int lj(long long u,long long v,long long w,long long f){
	top++;
	to[top]=v;
	liu[top]=w;
	fei[top]=f;
	nex[top]=fir[u];
	fir[u]=top;
	return 0;
}
int lj2(long long u,long long v,long long w,long long f){
	lj(u,v,w,f);
	lj(v,u,0,-f);
	return 0;
}
int main(){
	scanf("%lld",&n);
	s=(n<<1)+1;
	t=s+1;
	a=t+1;
	b=a+1;
	c=b+1;
	e=c+1;
	for(i=1;i<=n;i++)
		scanf("%lld%lld%lld",&rx[i],&ry[i],&rc[i]);
	for(i=1;i<=n;i++)
		scanf("%lld%lld%lld",&bx[i],&by[i],&bc[i]);
	for(i=1;i<=n;i++){
		lj2(i,a,rc[i],rx[i]+ry[i]);
		lj2(i,b,rc[i],rx[i]-ry[i]);
		lj2(i,c,rc[i],-rx[i]+ry[i]);
		lj2(i,e,rc[i],-rx[i]-ry[i]);
	}
	for(i=1;i<=n;i++){
		lj2(a,i+n,bc[i],-bx[i]-by[i]);
		lj2(b,i+n,bc[i],-bx[i]+by[i]);
		lj2(c,i+n,bc[i],bx[i]-by[i]);
		lj2(e,i+n,bc[i],bx[i]+by[i]);
	}
	for(i=1;i<=n;i++)
		lj2(s,i,rc[i],0);
	for(i=1;i<=n;i++)
		lj2(i+n,t,bc[i],0);
	n=e;
	o*=o;
	while(1)
	{
		head=1;tail=1;
		d[1]=s;
		
		for(i=1;i<=n;i++)
			vis[i]=-o;
		vis[s]=0;
		book[s]=1;
		while(head<=tail){
			top2=d[head];
			top1=fir[top2];
			while(top1!=0){
				if(vis[top2]+fei[top1]>vis[to[top1]]&&liu[top1]!=0){
					vis[to[top1]]=vis[top2]+fei[top1];
					zz[to[top1]]=top2;
					zz2[to[top1]]=liu[top1];
					zz3[to[top1]]=top1;
					if(book[to[top1]]==0)
					{
						tail++;
						d[tail]=to[top1];
						book[to[top1]]=1;
					}
				}
				top1=nex[top1];
			}
			head++;
			book[top2]=0;
		}
		if(vis[t]<0)
			break;
		top1=t;
		min1=o;
		while(top1!=s){
			min1=min(min1,zz2[top1]);
			top1=zz[top1];
		}
		top1=t;
		ans+=min1;
		ans2+=min1*vis[t];
		while(top1!=s){
			top2=zz3[top1];
			if(top2%2==1)
				top3=top2+1;
			else
				top3=top2-1;
			top1=zz[top1];
			liu[top2]-=min1;
			liu[top3]+=min1;
		}
	}
	cout<<ans2<<endl;
	return 0;
}
```