#include<bits/stdc++.h>
#define fo(i,l,r) for(int i=l;i<=r;i++)
#define of(i,l,r) for(int i=l;i>=r;i--)
#define fe(i,u) for(int i=head[u];i;i=e[i].next)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
inline void open(const char *s)
{
	#ifndef ONLINE_JUDGE
	char str[20];
	sprintf(str,"%s.in",s);
	freopen(str,"r",stdin);
//	sprintf(str,"%s.out",s);
//	freopen(str,"w",stdout);
	#endif
}
inline int rd()
{
	static int x,f;
	x=0;f=1;
	char ch=getchar();
	for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
	for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+ch-'0';
	return f>0?x:-x;
}
const int N=2010;
const int Inf=1e9;
int n,tot=1,head[N<<1],vs,vt;
struct edge{int v,f,w,next;}e[N*20];

inline void add(int u,int v,int f,int w)
{
	e[++tot]=edge{v,f,w,head[u]};head[u]=tot;
	e[++tot]=edge{u,0,-w,head[v]};head[v]=tot;
}

int cur[N<<1];ll d[N<<1];bool inq[N<<1];
inline bool bfs()
{static queue<int>q;
	while(!q.empty())q.pop();
	fo(i,1,vt)d[i]=-1e16,cur[i]=inq[i]=0;
	q.push(vs);d[vs]=0;cur[vs]=head[vs];inq[vs]=1;
	while(!q.empty()){
		int u=q.front();q.pop();inq[u]=0;
		fe(i,u){
			int v=e[i].v;
			if(e[i].f&&d[v]<d[u]+e[i].w){
				d[v]=d[u]+e[i].w;
				if(!inq[v]){
					cur[v]=head[v];
					q.push(v),inq[v]=1;
				}
			}
		}
	}
	return d[vt]!=-1e16;
}
int vis[N],tim=0;
int dfs(int u,int a)
{
	if(u==vt||!a)return a;
	int ans=0;vis[u]=tim;
	for(int&i=cur[u];i;i=e[i].next){
		int v=e[i].v,f=e[i].f;
		if(f&&vis[v]!=tim&&d[v]==d[u]+e[i].w){
			f=dfs(v,min(f,a));
			if(!f)continue;
			e[i].f-=f;e[i^1].f+=f;
			ans+=f;a-=f;if(!a)break;
		}
	}
	return ans;
}

int main()
{
	n=rd();vs=n+n+5;vt=vs+1;
	fo(i,1,n){
		int x=rd(),y=rd(),z=rd();
		add(vs,i,z,0);
		add(i,n+n+1,Inf,x+y);
		add(i,n+n+2,Inf,x-y);
		add(i,n+n+3,Inf,y-x);
		add(i,n+n+4,Inf,-x-y);
	}
	fo(i,1,n){
		int x=rd(),y=rd(),z=rd();
		x=-x;y=-y;
		add(i+n,vt,z,0);
		add(n+n+1,n+i,Inf,x+y);
		add(n+n+2,n+i,Inf,x-y);
		add(n+n+3,n+i,Inf,y-x);
		add(n+n+4,n+i,Inf,-x-y);
	}
	ll ans=0;
	while(bfs()){
//		while(1){
			++tim;
			int f=dfs(vs,Inf);
			ans+=d[vt]*f;
//			if(!f)break;
//		}
	}
	cout<<ans;
	return 0;
}

### 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<bits/stdc++.h>
#define fo(i,l,r) for(int i=l;i<=r;i++)
#define of(i,l,r) for(int i=l;i>=r;i--)
#define fe(i,u) for(int i=head[u];i;i=e[i].next)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
inline void open(const char *s)
{
	#ifndef ONLINE_JUDGE
	char str[20];
	sprintf(str,"%s.in",s);
	freopen(str,"r",stdin);
//	sprintf(str,"%s.out",s);
//	freopen(str,"w",stdout);
	#endif
}
inline int rd()
{
	static int x,f;
	x=0;f=1;
	char ch=getchar();
	for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
	for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+ch-'0';
	return f>0?x:-x;
}
const int N=2010;
const int Inf=1e9;
int n,tot=1,head[N<<1],vs,vt;
struct edge{int v,f,w,next;}e[N*20];

inline void add(int u,int v,int f,int w)
{
	e[++tot]=edge{v,f,w,head[u]};head[u]=tot;
	e[++tot]=edge{u,0,-w,head[v]};head[v]=tot;
}

int cur[N<<1];ll d[N<<1];bool inq[N<<1];
inline bool bfs()
{static queue<int>q;
	while(!q.empty())q.pop();
	fo(i,1,vt)d[i]=-1e16,cur[i]=inq[i]=0;
	q.push(vs);d[vs]=0;cur[vs]=head[vs];inq[vs]=1;
	while(!q.empty()){
		int u=q.front();q.pop();inq[u]=0;
		fe(i,u){
			int v=e[i].v;
			if(e[i].f&&d[v]<d[u]+e[i].w){
				d[v]=d[u]+e[i].w;
				if(!inq[v]){
					cur[v]=head[v];
					q.push(v),inq[v]=1;
				}
			}
		}
	}
	return d[vt]!=-1e16;
}
int vis[N],tim=0;
int dfs(int u,int a)
{
	if(u==vt||!a)return a;
	int ans=0;vis[u]=tim;
	for(int&i=cur[u];i;i=e[i].next){
		int v=e[i].v,f=e[i].f;
		if(f&&vis[v]!=tim&&d[v]==d[u]+e[i].w){
			f=dfs(v,min(f,a));
			if(!f)continue;
			e[i].f-=f;e[i^1].f+=f;
			ans+=f;a-=f;if(!a)break;
		}
	}
	return ans;
}

int main()
{
	n=rd();vs=n+n+5;vt=vs+1;
	fo(i,1,n){
		int x=rd(),y=rd(),z=rd();
		add(vs,i,z,0);
		add(i,n+n+1,Inf,x+y);
		add(i,n+n+2,Inf,x-y);
		add(i,n+n+3,Inf,y-x);
		add(i,n+n+4,Inf,-x-y);
	}
	fo(i,1,n){
		int x=rd(),y=rd(),z=rd();
		x=-x;y=-y;
		add(i+n,vt,z,0);
		add(n+n+1,n+i,Inf,x+y);
		add(n+n+2,n+i,Inf,x-y);
		add(n+n+3,n+i,Inf,y-x);
		add(n+n+4,n+i,Inf,-x-y);
	}
	ll ans=0;
	while(bfs()){
//		while(1){
			++tim;
			int f=dfs(vs,Inf);
			ans+=d[vt]*f;
//			if(!f)break;
//		}
	}
	cout<<ans;
	return 0;
}

```