#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <fstream>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <map>
#include <set>

using namespace std;

typedef long long ll;

const int MAXN = 2010;
const int MAXM = 40010;
const ll INF = 1ll << 60;

struct Edge
{
	int to;
	ll cap;
	ll val;
	int nxt;
}edge[MAXM];

int n,W,id = 1,S,T,tot;
int a[MAXN];
int first[MAXN];

ll dis[MAXN];

bool vis[MAXN];
bool inqueue[MAXN];

queue<int> Q;

pair<int,int> v[MAXN];

void addE(int u,int v,ll c,ll w)
{
	edge[++id] = (Edge){v,c,w,first[u]};
	first[u] = id;
	edge[++id] = (Edge){u,0,-w,first[v]};
	first[v] = id;
}

ll SPFA()
{
	for (int i = S;i <= T;i++)
		dis[i] = -INF;
	dis[S] = 0;
	Q.push(S);
	inqueue[S] = 1;
	while (!Q.empty())
	{
		int u = Q.front();
		Q.pop();
		inqueue[u] = 0;
		for (int i = first[u];i;i = edge[i].nxt)
			if (edge[i].cap > 0 && dis[edge[i].to] < dis[u] + edge[i].val)
			{
				dis[edge[i].to] = dis[u] + edge[i].val;
				if (!inqueue[edge[i].to])
				{
					Q.push(edge[i].to);
					inqueue[edge[i].to] = 1;
				}
			}
	}
	return dis[T];
}

ll dfs(int u,ll f,ll &tot)
{
	if (u == T)
		return f;
	vis[u] = 1;
	ll res = 0;
	for (int i = first[u];i;i = edge[i].nxt)
		if (!vis[edge[i].to] && edge[i].cap > 0 && dis[edge[i].to] == dis[u] + edge[i].val)
		{
			ll ret = dfs(edge[i].to,min(edge[i].cap,f),tot);
			edge[i].cap -= ret;
			edge[i ^ 1].cap += ret;
			f -= ret;
			res += ret;
			tot += ret * edge[i].val;
			if (!f)
				break;
		}
	return res;
}

ll mcmf()
{
	ll res = 0;
	while (SPFA() > 0)
	{
		memset(vis,0,sizeof(vis));
		dfs(S,INF,res);
	}
	return res;
}

int main()
{
	scanf("%d",&n);
	S = 0,T = 2 * n + 5;
	for (int a,b,c,i = 1;i <= n;i++)
	{
		scanf("%d%d%d",&b,&c,&a);
		addE(S,i,a,0);
		addE(i,2 * n + 1,INF,b + c);
		addE(i,2 * n + 2,INF,b - c);
		addE(i,2 * n + 3,INF,c - b);
		addE(i,2 * n + 4,INF,-b - c);
	}
	for (int a,b,c,i = 1;i <= n;i++)
	{
		scanf("%d%d%d",&b,&c,&a);
		addE(i + n,T,a,0);
		addE(2 * n + 1,i + n,INF,-b - c);
		addE(2 * n + 2,i + n,INF,c - b);
		addE(2 * n + 3,i + n,INF,b - c);
		addE(2 * n + 4,i + n,INF,b + c);
	}
	printf("%lld\n",mcmf());
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <fstream>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <map>
#include <set>

using namespace std;

typedef long long ll;

const int MAXN = 2010;
const int MAXM = 40010;
const ll INF = 1ll << 60;

struct Edge
{
	int to;
	ll cap;
	ll val;
	int nxt;
}edge[MAXM];

int n,W,id = 1,S,T,tot;
int a[MAXN];
int first[MAXN];

ll dis[MAXN];

bool vis[MAXN];
bool inqueue[MAXN];

queue<int> Q;

pair<int,int> v[MAXN];

void addE(int u,int v,ll c,ll w)
{
	edge[++id] = (Edge){v,c,w,first[u]};
	first[u] = id;
	edge[++id] = (Edge){u,0,-w,first[v]};
	first[v] = id;
}

ll SPFA()
{
	for (int i = S;i <= T;i++)
		dis[i] = -INF;
	dis[S] = 0;
	Q.push(S);
	inqueue[S] = 1;
	while (!Q.empty())
	{
		int u = Q.front();
		Q.pop();
		inqueue[u] = 0;
		for (int i = first[u];i;i = edge[i].nxt)
			if (edge[i].cap > 0 && dis[edge[i].to] < dis[u] + edge[i].val)
			{
				dis[edge[i].to] = dis[u] + edge[i].val;
				if (!inqueue[edge[i].to])
				{
					Q.push(edge[i].to);
					inqueue[edge[i].to] = 1;
				}
			}
	}
	return dis[T];
}

ll dfs(int u,ll f,ll &tot)
{
	if (u == T)
		return f;
	vis[u] = 1;
	ll res = 0;
	for (int i = first[u];i;i = edge[i].nxt)
		if (!vis[edge[i].to] && edge[i].cap > 0 && dis[edge[i].to] == dis[u] + edge[i].val)
		{
			ll ret = dfs(edge[i].to,min(edge[i].cap,f),tot);
			edge[i].cap -= ret;
			edge[i ^ 1].cap += ret;
			f -= ret;
			res += ret;
			tot += ret * edge[i].val;
			if (!f)
				break;
		}
	return res;
}

ll mcmf()
{
	ll res = 0;
	while (SPFA() > 0)
	{
		memset(vis,0,sizeof(vis));
		dfs(S,INF,res);
	}
	return res;
}

int main()
{
	scanf("%d",&n);
	S = 0,T = 2 * n + 5;
	for (int a,b,c,i = 1;i <= n;i++)
	{
		scanf("%d%d%d",&b,&c,&a);
		addE(S,i,a,0);
		addE(i,2 * n + 1,INF,b + c);
		addE(i,2 * n + 2,INF,b - c);
		addE(i,2 * n + 3,INF,c - b);
		addE(i,2 * n + 4,INF,-b - c);
	}
	for (int a,b,c,i = 1;i <= n;i++)
	{
		scanf("%d%d%d",&b,&c,&a);
		addE(i + n,T,a,0);
		addE(2 * n + 1,i + n,INF,-b - c);
		addE(2 * n + 2,i + n,INF,c - b);
		addE(2 * n + 3,i + n,INF,b - c);
		addE(2 * n + 4,i + n,INF,b + c);
	}
	printf("%lld\n",mcmf());
	return 0;
}
```