#include<bits/stdc++.h>
#define ll long long
using namespace std;
template<typename tn> void read(tn &a){
	tn x=0,f=1; char c=' ';
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	a=x*f;
}
const int N = 101000;
struct edge{int v,nxt,f;ll w;}e[N];
int f[N],fr[N],in[N],n,tot,t,flow,head[N];
ll d[N],ans;
void add(int x,int y,int f,ll w){
	e[++tot]={y,head[x],f,w};
	head[x]=tot;
	e[++tot]={x,head[y],0,-w};
	head[y]=tot;
}
bool spfa(){
	memset(d,192,sizeof(d));d[0]=0;
	memset(f,0,sizeof(f));f[0]=1e9;
	memset(fr,0,sizeof(fr));
	queue<int> q;q.push(0);
	while(!q.empty()){
		int u=q.front();q.pop();in[u]=0;
		for(int i=head[u];i;i=e[i].nxt)
			if(e[i].f&&d[e[i].v]<d[u]+e[i].w){
				d[e[i].v]=d[u]+e[i].w;
				f[e[i].v]=min(f[u],e[i].f);
				fr[e[i].v]=i;
				if(!in[e[i].v]) q.push(e[i].v),in[e[i].v]=1;
			}
	}
	return f[t];
}
void update(){
	flow+=f[t];ans+=f[t]*d[t];
	for(int x=t;fr[x];x=e[fr[x]^1].v)
		e[fr[x]].f-=f[t],e[fr[x]^1].f+=f[t];
}
ll EK(){
	ans=flow=0;
	while(spfa()) update();
	return ans;
}
int main(){
	read(n);tot=1;t=2*n+5;
	for(int i=1;i<=n;i++){
		int x,y,z;read(x);read(y);read(z);
		add(0,i,z,0);
		add(i,2*n+1,z,x+y);
		add(i,2*n+2,z,x-y);
		add(i,2*n+3,z,-x+y);
		add(i,2*n+4,z,-x-y);
	}
	for(int i=1;i<=n;i++){
		int x,y,z;read(x);read(y);read(z);
		add(i+n,t,z,0);
		add(2*n+4,i+n,z,x+y);
		add(2*n+3,i+n,z,x-y);
		add(2*n+1,i+n,z,-x-y);
		add(2*n+2,i+n,z,-x+y);
	}
	cout<<EK()<<'\n';
	return 0;
}

### Prompt

Please provide a Cpp coded solution 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<bits/stdc++.h>
#define ll long long
using namespace std;
template<typename tn> void read(tn &a){
	tn x=0,f=1; char c=' ';
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	a=x*f;
}
const int N = 101000;
struct edge{int v,nxt,f;ll w;}e[N];
int f[N],fr[N],in[N],n,tot,t,flow,head[N];
ll d[N],ans;
void add(int x,int y,int f,ll w){
	e[++tot]={y,head[x],f,w};
	head[x]=tot;
	e[++tot]={x,head[y],0,-w};
	head[y]=tot;
}
bool spfa(){
	memset(d,192,sizeof(d));d[0]=0;
	memset(f,0,sizeof(f));f[0]=1e9;
	memset(fr,0,sizeof(fr));
	queue<int> q;q.push(0);
	while(!q.empty()){
		int u=q.front();q.pop();in[u]=0;
		for(int i=head[u];i;i=e[i].nxt)
			if(e[i].f&&d[e[i].v]<d[u]+e[i].w){
				d[e[i].v]=d[u]+e[i].w;
				f[e[i].v]=min(f[u],e[i].f);
				fr[e[i].v]=i;
				if(!in[e[i].v]) q.push(e[i].v),in[e[i].v]=1;
			}
	}
	return f[t];
}
void update(){
	flow+=f[t];ans+=f[t]*d[t];
	for(int x=t;fr[x];x=e[fr[x]^1].v)
		e[fr[x]].f-=f[t],e[fr[x]^1].f+=f[t];
}
ll EK(){
	ans=flow=0;
	while(spfa()) update();
	return ans;
}
int main(){
	read(n);tot=1;t=2*n+5;
	for(int i=1;i<=n;i++){
		int x,y,z;read(x);read(y);read(z);
		add(0,i,z,0);
		add(i,2*n+1,z,x+y);
		add(i,2*n+2,z,x-y);
		add(i,2*n+3,z,-x+y);
		add(i,2*n+4,z,-x-y);
	}
	for(int i=1;i<=n;i++){
		int x,y,z;read(x);read(y);read(z);
		add(i+n,t,z,0);
		add(2*n+4,i+n,z,x+y);
		add(2*n+3,i+n,z,x-y);
		add(2*n+1,i+n,z,-x-y);
		add(2*n+2,i+n,z,-x+y);
	}
	cout<<EK()<<'\n';
	return 0;
}

```