#include<iostream>
#include<cstring>
#include<cassert>
#include<cmath>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<time.h>
#include<algorithm>
using namespace std;
#define REP(i,x,y) for(ll i=x;i<=y;i++)
#define rep(i,n) REP(i,1,n)
#define rep0(i,n) REP(i,0,n-1)
#define repG(i,x) for(ll i=pos[x];~i;i=e[i].next)
#define ll long long
#define db double
const ll N=1e4+7;
const ll INF=1e15+7;
ll pos[N],dis[N],Qu[N*90],cur[N];
bool vis[N];
struct edge{ll v,next,c,f;}e[N*2];
ll n,num=0,S,T,ans=0;
void add(ll x,ll y,ll c,ll f){
	e[num]=(edge){y,pos[x],c,f}; pos[x]=num++;
	e[num]=(edge){x,pos[y],0,-f}; pos[y]=num++;
}
bool spfa(){
	ll hd=0,ti=1;
	Qu[1]=S;
	rep(i,n+n+10){
		dis[i]=INF;
		vis[i]=1;
		cur[i]=pos[i];
	}
	dis[S]=0;
	while(hd!=ti){
		ll t=Qu[++hd];
		vis[t]=1;
		repG(i,t){
			if(dis[e[i].v]>dis[t]+e[i].f&&e[i].c){
				dis[e[i].v]=dis[t]+e[i].f;
				if(vis[e[i].v]){
					vis[e[i].v]=0;
					Qu[++ti]=e[i].v;
				}
			}
		}
	}
	return dis[T]<INF;
}
ll dfs(ll x,ll f){
	if(x==T){
		ans+=f*dis[T];
		return f;
	}
	vis[x]=0;
	ll used=0;
	for(;~cur[x];cur[x]=e[cur[x]].next){
		ll i=cur[x];
		if(e[i].c&&dis[e[i].v]==dis[x]+e[i].f&&vis[e[i].v]){
			ll k=dfs(e[i].v,min(e[i].c,f-used));
			used+=k;
			e[i].c-=k;
			e[i^1].c+=k;
		}
		if(used==f)break;
	}
	if(used==f)vis[x]=1;
	return used;
}
int main(){
	memset(pos,-1,sizeof(pos));
	scanf("%lld",&n);
	S=2*n+1;
	T=2*n+2;
	REP(i,2,5)add(n+n+2*i-1,n+n+2*i,INF,0);
	rep(i,n){
		ll c,x,y; scanf("%lld%lld%lld",&x,&y,&c);
		add(S,i,c,0);
		add(i,n+n+3,c,-x-y);
		add(i,n+n+5,c,-x+y);
		add(i,n+n+7,c,x-y);
		add(i,n+n+9,c,x+y);
	}
	rep(i,n){
		ll c,x,y; scanf("%lld%lld%lld",&x,&y,&c);
		add(i+n,T,c,0);
		add(n+n+4,i+n,c,x+y);
		add(n+n+6,i+n,c,x-y);
		add(n+n+8,i+n,c,-x+y);
		add(n+n+10,i+n,c,-x-y);
	}
	while(spfa())dfs(S,INF);
	printf("%lld\n",-ans);
	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<iostream>
#include<cstring>
#include<cassert>
#include<cmath>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<time.h>
#include<algorithm>
using namespace std;
#define REP(i,x,y) for(ll i=x;i<=y;i++)
#define rep(i,n) REP(i,1,n)
#define rep0(i,n) REP(i,0,n-1)
#define repG(i,x) for(ll i=pos[x];~i;i=e[i].next)
#define ll long long
#define db double
const ll N=1e4+7;
const ll INF=1e15+7;
ll pos[N],dis[N],Qu[N*90],cur[N];
bool vis[N];
struct edge{ll v,next,c,f;}e[N*2];
ll n,num=0,S,T,ans=0;
void add(ll x,ll y,ll c,ll f){
	e[num]=(edge){y,pos[x],c,f}; pos[x]=num++;
	e[num]=(edge){x,pos[y],0,-f}; pos[y]=num++;
}
bool spfa(){
	ll hd=0,ti=1;
	Qu[1]=S;
	rep(i,n+n+10){
		dis[i]=INF;
		vis[i]=1;
		cur[i]=pos[i];
	}
	dis[S]=0;
	while(hd!=ti){
		ll t=Qu[++hd];
		vis[t]=1;
		repG(i,t){
			if(dis[e[i].v]>dis[t]+e[i].f&&e[i].c){
				dis[e[i].v]=dis[t]+e[i].f;
				if(vis[e[i].v]){
					vis[e[i].v]=0;
					Qu[++ti]=e[i].v;
				}
			}
		}
	}
	return dis[T]<INF;
}
ll dfs(ll x,ll f){
	if(x==T){
		ans+=f*dis[T];
		return f;
	}
	vis[x]=0;
	ll used=0;
	for(;~cur[x];cur[x]=e[cur[x]].next){
		ll i=cur[x];
		if(e[i].c&&dis[e[i].v]==dis[x]+e[i].f&&vis[e[i].v]){
			ll k=dfs(e[i].v,min(e[i].c,f-used));
			used+=k;
			e[i].c-=k;
			e[i^1].c+=k;
		}
		if(used==f)break;
	}
	if(used==f)vis[x]=1;
	return used;
}
int main(){
	memset(pos,-1,sizeof(pos));
	scanf("%lld",&n);
	S=2*n+1;
	T=2*n+2;
	REP(i,2,5)add(n+n+2*i-1,n+n+2*i,INF,0);
	rep(i,n){
		ll c,x,y; scanf("%lld%lld%lld",&x,&y,&c);
		add(S,i,c,0);
		add(i,n+n+3,c,-x-y);
		add(i,n+n+5,c,-x+y);
		add(i,n+n+7,c,x-y);
		add(i,n+n+9,c,x+y);
	}
	rep(i,n){
		ll c,x,y; scanf("%lld%lld%lld",&x,&y,&c);
		add(i+n,T,c,0);
		add(n+n+4,i+n,c,x+y);
		add(n+n+6,i+n,c,x-y);
		add(n+n+8,i+n,c,-x+y);
		add(n+n+10,i+n,c,-x-y);
	}
	while(spfa())dfs(S,INF);
	printf("%lld\n",-ans);
	return 0;
}

```