#include<cstdio>
#include<cstring>
#include<queue>
#define inf 0x3f3f3f3f

using namespace std;

const int N = 2010;

struct note
{
	int to , nex , val , flow;
}g[N << 4];

int n , m , st[N] , S , T , vis[N];
long long ans , dist[N];


inline void add(int u , int v , int val , int flow)
{
	static int tot = 1;
	g[++ tot] = (note){v , st[u] , val , flow} , st[u] = tot;
}

inline void add_edge(int u , int v , int flow ,int val){add(u , v , val , flow) , add(v , u , -val , 0);}

queue<int> q;

int spfa()
{
	while(!q.empty()) q.pop();q.push(S);
	memset(vis , 0 , sizeof vis) , vis[S] = 1;
	memset(dist , -0x3f , sizeof dist) , dist[S] = 0;

	while(!q.empty())
	{
		int u = q.front();q.pop() , vis[u] = 0;
		for(int i = st[u] ; i ;i = g[i].nex)
		{
			int v = g[i].to;
			if(g[i].flow && dist[v] < dist[u] + g[i].val)
				dist[v] = dist[u] + g[i].val , (!vis[v]) && (q.push(v) , vis[v] = 1);
		}
	}

	return dist[T] > -inf;
}

int dfs(int now , int flow)
{
	if(now == T) return flow;
	vis[now] = 1;int ret = 0 , tmp;

	for(int i = st[now] ; i ; i = g[i].nex)
	{
		int v = g[i].to;
		if(vis[v] || !g[i].flow || dist[v] != dist[now] + g[i].val) continue;

		tmp = dfs(v , min(g[i].flow , flow));
		
		if(tmp)
			g[i].flow -= tmp , g[i ^ 1].flow += tmp , flow -= tmp , ret += tmp , ans += 1ll * tmp * g[i].val;
		else
			dist[v] = -inf;
	}

	return ret;
}


int main()
{
	scanf("%d",&n) , S = n << 1 | 1, T = S + 1;

	for(int i = 1 , x , y , t;i <= n;i ++) 
		scanf("%d %d %d",&x,&y,&t) , add_edge(S , i , t , 0) , 
		add_edge(i , T + 1 , t , x + y) , add_edge(i , T + 2 , t , x - y) , 
		add_edge(i , T + 3 , t , -x + y) , add_edge(i , T + 4 , t , -x -y) ;
	for(int i = 1 , x , y , t;i <= n;i ++)
		scanf("%d %d %d",&x,&y,&t) , add_edge(i + n, T , t , 0) , 
		add_edge(T + 1 , i + n , t , -x - y) , add_edge(T + 2 , i + n , t , -x + y) , 
		add_edge(T + 3 , i + n , t , x - y) , add_edge(T + 4 , i + n , t , +x + y);

	while(spfa()) dfs(S , inf);
	printf("%lld\n",ans);

	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<cstdio>
#include<cstring>
#include<queue>
#define inf 0x3f3f3f3f

using namespace std;

const int N = 2010;

struct note
{
	int to , nex , val , flow;
}g[N << 4];

int n , m , st[N] , S , T , vis[N];
long long ans , dist[N];


inline void add(int u , int v , int val , int flow)
{
	static int tot = 1;
	g[++ tot] = (note){v , st[u] , val , flow} , st[u] = tot;
}

inline void add_edge(int u , int v , int flow ,int val){add(u , v , val , flow) , add(v , u , -val , 0);}

queue<int> q;

int spfa()
{
	while(!q.empty()) q.pop();q.push(S);
	memset(vis , 0 , sizeof vis) , vis[S] = 1;
	memset(dist , -0x3f , sizeof dist) , dist[S] = 0;

	while(!q.empty())
	{
		int u = q.front();q.pop() , vis[u] = 0;
		for(int i = st[u] ; i ;i = g[i].nex)
		{
			int v = g[i].to;
			if(g[i].flow && dist[v] < dist[u] + g[i].val)
				dist[v] = dist[u] + g[i].val , (!vis[v]) && (q.push(v) , vis[v] = 1);
		}
	}

	return dist[T] > -inf;
}

int dfs(int now , int flow)
{
	if(now == T) return flow;
	vis[now] = 1;int ret = 0 , tmp;

	for(int i = st[now] ; i ; i = g[i].nex)
	{
		int v = g[i].to;
		if(vis[v] || !g[i].flow || dist[v] != dist[now] + g[i].val) continue;

		tmp = dfs(v , min(g[i].flow , flow));
		
		if(tmp)
			g[i].flow -= tmp , g[i ^ 1].flow += tmp , flow -= tmp , ret += tmp , ans += 1ll * tmp * g[i].val;
		else
			dist[v] = -inf;
	}

	return ret;
}


int main()
{
	scanf("%d",&n) , S = n << 1 | 1, T = S + 1;

	for(int i = 1 , x , y , t;i <= n;i ++) 
		scanf("%d %d %d",&x,&y,&t) , add_edge(S , i , t , 0) , 
		add_edge(i , T + 1 , t , x + y) , add_edge(i , T + 2 , t , x - y) , 
		add_edge(i , T + 3 , t , -x + y) , add_edge(i , T + 4 , t , -x -y) ;
	for(int i = 1 , x , y , t;i <= n;i ++)
		scanf("%d %d %d",&x,&y,&t) , add_edge(i + n, T , t , 0) , 
		add_edge(T + 1 , i + n , t , -x - y) , add_edge(T + 2 , i + n , t , -x + y) , 
		add_edge(T + 3 , i + n , t , x - y) , add_edge(T + 4 , i + n , t , +x + y);

	while(spfa()) dfs(S , inf);
	printf("%lld\n",ans);

	return 0;
}

```