#include<bits/stdc++.h>

using namespace std;

#define N 1000
#define ll long long
#define dt 2000000000
#define inf 0x3f3f3f3f3f3f3f3f

ll n,cnt = 1,S,T,cm,fs;
ll head[2*N+10],last[2*N+10],cost[2*N+10],flow[2*N+10];
ll rx[2*N+10],rc[2*N+10],ry[2*N+10],bx[2*N+10],bc[2*N+10],by[2*N+10];
bool inq[2*N+10];
queue<ll>q;

struct edge{
	ll des,next,flow,cost;
}e[30005];

void addedge(ll a,ll b,ll f,ll c){
	cnt++;
	e[cnt].des = b;
	e[cnt].cost = c;
	e[cnt].flow = f;
	e[cnt].next = head[a];
	head[a] = cnt;
}

void add(ll a,ll b,ll f,ll c){
	addedge(a,b,f,c);
	addedge(b,a,0,-c);
}

inline ll read(){
	ll s = 0,w = 1;
	char ch = getchar();
	while(!isdigit(ch)){if(ch == '-')w = -1;ch = getchar();}
	while(isdigit(ch)){s = 10 * s + ch - '0';ch = getchar();}
	return s * w;
}

bool spfa(){
	ll i,u,v;
	memset(cost,0x3f,sizeof(cost));
	inq[S] = 1;
	cost[S] = 0;
	flow[S] = inf;
	q.push(S);
	while(!q.empty()){
		u = q.front();
		q.pop();
		inq[u] = 0;
		for(i = head[u];i;i = e[i].next){
			if(!e[i].flow)continue;
			v = e[i].des;
			if(cost[u] + e[i].cost < cost[v]){
				cost[v] = cost[u] + e[i].cost;
				last[v] = i;
				flow[v] = min(flow[u],e[i].flow);
				if(!inq[v]){
					q.push(v);
					inq[v] = 1;
				}
			}
		}
	}
	return cost[T] != inf;
}

void mcmf(){
	ll df,dc,u;
	while(spfa()){
		df = flow[T];
		dc = cost[T];
		fs += df,cm += df * dc;
		for(u = T;u != S;u = e[last[u] ^ 1].des){
			e[last[u]].flow -= df;
			e[last[u] ^ 1].flow += df;
		}
	}
}

int main(){
//	freopen("D.in","r",stdin);
//	freopen("D.out","w",stdout);
	n = read();
	ll i,j;
	ll s = 0;
	for(i = 1;i <= n;i++){
		rx[i] = read();ry[i] = read();rc[i] = read();
	}
	for(i = 1;i <= n;i++){
		bx[i] = read();by[i] = read();bc[i] = read();
	}
	S = 0,T = 2 * n + 5;
	for(i = 1;i <= n;i++)add(S,i,rc[i],0);
	for(i = 1;i <= n;i++)add(i+n+4,T,bc[i],0);
	for(i = 1;i <= n;i++){
		add(i,n+1,inf,rx[i]+ry[i]+dt);
		add(i,n+2,inf,rx[i]-ry[i]+dt);
		add(i,n+3,inf,-rx[i]+ry[i]+dt);
		add(i,n+4,inf,-rx[i]-ry[i]+dt);
		add(n+1,i+n+4,inf,-bx[i]-by[i]+dt);
		add(n+2,i+n+4,inf,-bx[i]+by[i]+dt);
		add(n+3,i+n+4,inf,bx[i]-by[i]+dt);
		add(n+4,i+n+4,inf,bx[i]+by[i]+dt);
	}
	mcmf();
/*for(int u = S;u <= T;u++){
cout<<"u="<<u<<endl;
	for(i = head[u];i;i = e[i].next){
		int v = e[i].des;
		cout<<"v="<<v<<" cost="<<e[i].cost<<" flow="<<e[i].flow<<endl;
	}
} */
//cout<<"cm="<<cm<<" fs="<<fs<<endl;
	cout << fs * 2 * dt - cm;
//	fclose(stdin);
//	fclose(stdout);
	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>

using namespace std;

#define N 1000
#define ll long long
#define dt 2000000000
#define inf 0x3f3f3f3f3f3f3f3f

ll n,cnt = 1,S,T,cm,fs;
ll head[2*N+10],last[2*N+10],cost[2*N+10],flow[2*N+10];
ll rx[2*N+10],rc[2*N+10],ry[2*N+10],bx[2*N+10],bc[2*N+10],by[2*N+10];
bool inq[2*N+10];
queue<ll>q;

struct edge{
	ll des,next,flow,cost;
}e[30005];

void addedge(ll a,ll b,ll f,ll c){
	cnt++;
	e[cnt].des = b;
	e[cnt].cost = c;
	e[cnt].flow = f;
	e[cnt].next = head[a];
	head[a] = cnt;
}

void add(ll a,ll b,ll f,ll c){
	addedge(a,b,f,c);
	addedge(b,a,0,-c);
}

inline ll read(){
	ll s = 0,w = 1;
	char ch = getchar();
	while(!isdigit(ch)){if(ch == '-')w = -1;ch = getchar();}
	while(isdigit(ch)){s = 10 * s + ch - '0';ch = getchar();}
	return s * w;
}

bool spfa(){
	ll i,u,v;
	memset(cost,0x3f,sizeof(cost));
	inq[S] = 1;
	cost[S] = 0;
	flow[S] = inf;
	q.push(S);
	while(!q.empty()){
		u = q.front();
		q.pop();
		inq[u] = 0;
		for(i = head[u];i;i = e[i].next){
			if(!e[i].flow)continue;
			v = e[i].des;
			if(cost[u] + e[i].cost < cost[v]){
				cost[v] = cost[u] + e[i].cost;
				last[v] = i;
				flow[v] = min(flow[u],e[i].flow);
				if(!inq[v]){
					q.push(v);
					inq[v] = 1;
				}
			}
		}
	}
	return cost[T] != inf;
}

void mcmf(){
	ll df,dc,u;
	while(spfa()){
		df = flow[T];
		dc = cost[T];
		fs += df,cm += df * dc;
		for(u = T;u != S;u = e[last[u] ^ 1].des){
			e[last[u]].flow -= df;
			e[last[u] ^ 1].flow += df;
		}
	}
}

int main(){
//	freopen("D.in","r",stdin);
//	freopen("D.out","w",stdout);
	n = read();
	ll i,j;
	ll s = 0;
	for(i = 1;i <= n;i++){
		rx[i] = read();ry[i] = read();rc[i] = read();
	}
	for(i = 1;i <= n;i++){
		bx[i] = read();by[i] = read();bc[i] = read();
	}
	S = 0,T = 2 * n + 5;
	for(i = 1;i <= n;i++)add(S,i,rc[i],0);
	for(i = 1;i <= n;i++)add(i+n+4,T,bc[i],0);
	for(i = 1;i <= n;i++){
		add(i,n+1,inf,rx[i]+ry[i]+dt);
		add(i,n+2,inf,rx[i]-ry[i]+dt);
		add(i,n+3,inf,-rx[i]+ry[i]+dt);
		add(i,n+4,inf,-rx[i]-ry[i]+dt);
		add(n+1,i+n+4,inf,-bx[i]-by[i]+dt);
		add(n+2,i+n+4,inf,-bx[i]+by[i]+dt);
		add(n+3,i+n+4,inf,bx[i]-by[i]+dt);
		add(n+4,i+n+4,inf,bx[i]+by[i]+dt);
	}
	mcmf();
/*for(int u = S;u <= T;u++){
cout<<"u="<<u<<endl;
	for(i = head[u];i;i = e[i].next){
		int v = e[i].des;
		cout<<"v="<<v<<" cost="<<e[i].cost<<" flow="<<e[i].flow<<endl;
	}
} */
//cout<<"cm="<<cm<<" fs="<<fs<<endl;
	cout << fs * 2 * dt - cm;
//	fclose(stdin);
//	fclose(stdout);
	return 0;
}
```