#include <iostream>
#include <cstdio>
#include <queue>
#define int long long
#define ll long long
using namespace std;
const int N = 2105;
const int inf = 1e18;
queue<int> q;
int n, s, t, p1, p2, p3, p4, head[N], tot = 1, pre[N], last[N], vis[N], flow[N];
ll dis[N];
struct node{int to, nex, val, cost;}a[500005];
inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
	return x * f;
}
void add(int x, int y, int z, int w)
{
	a[++ tot].to = y; a[tot].nex = head[x]; a[tot].val = z; a[tot].cost = w; head[x] = tot;
	a[++ tot].to = x; a[tot].nex = head[y]; a[tot].val = 0; a[tot].cost = -w; head[y] = tot;
}
int spfa()
{
	for(int i = s; i <= p4; i ++) dis[i] = - inf, flow[i] = inf;
	dis[s] = 0; q.push(s); vis[s] = 1; pre[t] = -1;
	while(!q.empty())
	{
		int x = q.front(); q.pop(); vis[x] = 0;
		for(int i = head[x]; i; i = a[i].nex)
		{
			int y = a[i].to;
			if(dis[y] < dis[x] + a[i].cost && a[i].val)
			{
				dis[y] = dis[x] + a[i].cost; last[y] = i;
				pre[y] = x; flow[y] = min(flow[x], a[i].val);
				if(!vis[y]) vis[y] = 1, q.push(y);
			}
		}
	}
	return pre[t] != -1;
}
ll mcmf()
{
	ll mxcost = 0;
	while(spfa())
	{
		mxcost += flow[t] * dis[t]; int now = t;
		while(now != s)
		{
			a[last[now]].val -= flow[t];
			a[last[now] ^ 1].val += flow[t];
			now = pre[now];
		}
	}
	return mxcost;
}
signed main()
{
//	freopen(".in", "r", stdin);
//	freopen(".out", "w", stdout);
	n = read(); s = 0; t = 2 * n + 1;
	p1 = t + 1; p2 = p1 + 1; p3 = p2 + 1; p4 = p3 + 1;
	for(int i = 1, x, y, z; i <= n; i ++)
	{
		x = read(); y = read(); z = read(); add(s, i, z, 0);
		add(i, p1, inf, x + y); add(i, p2, inf, x - y);
		add(i, p3, inf, y - x); add(i, p4, inf, - x - y);
	}
	for(int i = 1, x, y, z; i <= n; i ++)
	{
		x = read(); y = read(); z = read(); add(i + n, t, z, 0);
		add(p1, i + n, inf, - x - y); add(p2, i + n, inf, y - x);
		add(p3, i + n, inf, x - y); add(p4, i + n, inf, x + y);
	}
	printf("%lld\n", mcmf());
	fclose(stdin);
	fclose(stdout);
	return 0;
}

### Prompt

Create a solution in cpp for 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 <cstdio>
#include <queue>
#define int long long
#define ll long long
using namespace std;
const int N = 2105;
const int inf = 1e18;
queue<int> q;
int n, s, t, p1, p2, p3, p4, head[N], tot = 1, pre[N], last[N], vis[N], flow[N];
ll dis[N];
struct node{int to, nex, val, cost;}a[500005];
inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
	return x * f;
}
void add(int x, int y, int z, int w)
{
	a[++ tot].to = y; a[tot].nex = head[x]; a[tot].val = z; a[tot].cost = w; head[x] = tot;
	a[++ tot].to = x; a[tot].nex = head[y]; a[tot].val = 0; a[tot].cost = -w; head[y] = tot;
}
int spfa()
{
	for(int i = s; i <= p4; i ++) dis[i] = - inf, flow[i] = inf;
	dis[s] = 0; q.push(s); vis[s] = 1; pre[t] = -1;
	while(!q.empty())
	{
		int x = q.front(); q.pop(); vis[x] = 0;
		for(int i = head[x]; i; i = a[i].nex)
		{
			int y = a[i].to;
			if(dis[y] < dis[x] + a[i].cost && a[i].val)
			{
				dis[y] = dis[x] + a[i].cost; last[y] = i;
				pre[y] = x; flow[y] = min(flow[x], a[i].val);
				if(!vis[y]) vis[y] = 1, q.push(y);
			}
		}
	}
	return pre[t] != -1;
}
ll mcmf()
{
	ll mxcost = 0;
	while(spfa())
	{
		mxcost += flow[t] * dis[t]; int now = t;
		while(now != s)
		{
			a[last[now]].val -= flow[t];
			a[last[now] ^ 1].val += flow[t];
			now = pre[now];
		}
	}
	return mxcost;
}
signed main()
{
//	freopen(".in", "r", stdin);
//	freopen(".out", "w", stdout);
	n = read(); s = 0; t = 2 * n + 1;
	p1 = t + 1; p2 = p1 + 1; p3 = p2 + 1; p4 = p3 + 1;
	for(int i = 1, x, y, z; i <= n; i ++)
	{
		x = read(); y = read(); z = read(); add(s, i, z, 0);
		add(i, p1, inf, x + y); add(i, p2, inf, x - y);
		add(i, p3, inf, y - x); add(i, p4, inf, - x - y);
	}
	for(int i = 1, x, y, z; i <= n; i ++)
	{
		x = read(); y = read(); z = read(); add(i + n, t, z, 0);
		add(p1, i + n, inf, - x - y); add(p2, i + n, inf, y - x);
		add(p3, i + n, inf, x - y); add(p4, i + n, inf, x + y);
	}
	printf("%lld\n", mcmf());
	fclose(stdin);
	fclose(stdout);
	return 0;
}

```