#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<string>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<set>
#include<map>
using namespace std;
typedef long long LL;
#define int long long
const int N = 1010 , inf = 1e16;
inline int read()
{
	register int x = 0 , f = 0; register char c = getchar();
	while(c < '0' || c > '9') f |= c == '-' , c = getchar();
	while(c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0' , c = getchar();
	return f ? -x : x;
}

int n , S , T , cnt = 1;
int head[N << 1] , f[N << 1] , pre[N << 1] , vis[N << 1];
LL dis[N << 1];
struct node{ int x , y , c; }p1[N] , p2[N];
struct edge{ int v , nex , c; LL val; }e[(N * (N + 2)) << 1];
inline void add(int u , int v , int c , int val)
{
	e[++cnt].v = v; e[cnt].nex = head[u]; e[cnt].c = c; e[cnt].val = val; head[u] = cnt;
	e[++cnt].v = u; e[cnt].nex = head[v]; e[cnt].c = 0; e[cnt].val = -val; head[v] = cnt;
}

queue<int> q;
bool spfa()
{
	for(int i = 1 ; i <= T ; ++i) dis[i] = -inf , pre[i] = 0 , f[i] = 0; dis[S] = 0; f[S] = inf; q.push(S);
	while(q.size())
	{
		int x = q.front(); q.pop(); vis[x] = 0;
		for(int i = head[x] , v; i ; i = e[i].nex)
		{
			v = e[i].v;
			if(e[i].c && dis[v] < dis[x] + e[i].val)
			{
				dis[v] = dis[x] + e[i].val; pre[v] = i; f[v] = min(f[x] , e[i].c);
				if(!vis[v]) vis[v] = 1 , q.push(v);
			}
		}
	}
	return dis[T] != -inf;
}

LL calc()
{
	LL cost = 0;
	while(spfa())
	{
		cost += (LL)f[T] * dis[T]; int t = T , i;
		do i = pre[t] , e[i].c -= f[T] , e[i^1].c += f[T] , t = e[i^1].v; while(t != S);
	}
	return cost;
}

inline int Dis(node a , node b) { return abs(a.x - b.x) + abs(a.y - b.y); }

signed main()
{
	n = read(); int t1 = n + n + 1 , t2 = t1 + 1 , t3 = t2 + 1 , t4 = t3 + 1; S = t4 + 1; T = S + 1;
	for(int i = 1 ; i <= n ; ++i) p1[i].x = read() , p1[i].y = read() , p1[i].c = read();
	for(int i = 1 ; i <= n ; ++i) p2[i].x = read() , p2[i].y = read() , p2[i].c = read();
	for(int i = 1 ; i <= n ; ++i)
	{
		add(S , i , p1[i].c , 0);
		add(i , t1 , inf , p1[i].x + p1[i].y);
		add(i , t2 , inf , p1[i].x - p1[i].y);
		add(i , t3 , inf , -p1[i].x + p1[i].y);
		add(i , t4 , inf , -p1[i].x - p1[i].y);
	}
	
	for(int i = 1 ; i <= n ; ++i)
	{
		add(i + n , T , p2[i].c , 0);
		add(t1 , i + n , inf , -p2[i].x - p2[i].y);
		add(t2 , i + n , inf , -p2[i].x + p2[i].y);
		add(t3 , i + n , inf , p2[i].x - p2[i].y);
		add(t4 , i + n , inf , p2[i].x + p2[i].y);
	}
	
	cout << calc() << '\n'; return 0;
}
### Prompt

Please create a solution in CPP 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<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<string>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<set>
#include<map>
using namespace std;
typedef long long LL;
#define int long long
const int N = 1010 , inf = 1e16;
inline int read()
{
	register int x = 0 , f = 0; register char c = getchar();
	while(c < '0' || c > '9') f |= c == '-' , c = getchar();
	while(c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0' , c = getchar();
	return f ? -x : x;
}

int n , S , T , cnt = 1;
int head[N << 1] , f[N << 1] , pre[N << 1] , vis[N << 1];
LL dis[N << 1];
struct node{ int x , y , c; }p1[N] , p2[N];
struct edge{ int v , nex , c; LL val; }e[(N * (N + 2)) << 1];
inline void add(int u , int v , int c , int val)
{
	e[++cnt].v = v; e[cnt].nex = head[u]; e[cnt].c = c; e[cnt].val = val; head[u] = cnt;
	e[++cnt].v = u; e[cnt].nex = head[v]; e[cnt].c = 0; e[cnt].val = -val; head[v] = cnt;
}

queue<int> q;
bool spfa()
{
	for(int i = 1 ; i <= T ; ++i) dis[i] = -inf , pre[i] = 0 , f[i] = 0; dis[S] = 0; f[S] = inf; q.push(S);
	while(q.size())
	{
		int x = q.front(); q.pop(); vis[x] = 0;
		for(int i = head[x] , v; i ; i = e[i].nex)
		{
			v = e[i].v;
			if(e[i].c && dis[v] < dis[x] + e[i].val)
			{
				dis[v] = dis[x] + e[i].val; pre[v] = i; f[v] = min(f[x] , e[i].c);
				if(!vis[v]) vis[v] = 1 , q.push(v);
			}
		}
	}
	return dis[T] != -inf;
}

LL calc()
{
	LL cost = 0;
	while(spfa())
	{
		cost += (LL)f[T] * dis[T]; int t = T , i;
		do i = pre[t] , e[i].c -= f[T] , e[i^1].c += f[T] , t = e[i^1].v; while(t != S);
	}
	return cost;
}

inline int Dis(node a , node b) { return abs(a.x - b.x) + abs(a.y - b.y); }

signed main()
{
	n = read(); int t1 = n + n + 1 , t2 = t1 + 1 , t3 = t2 + 1 , t4 = t3 + 1; S = t4 + 1; T = S + 1;
	for(int i = 1 ; i <= n ; ++i) p1[i].x = read() , p1[i].y = read() , p1[i].c = read();
	for(int i = 1 ; i <= n ; ++i) p2[i].x = read() , p2[i].y = read() , p2[i].c = read();
	for(int i = 1 ; i <= n ; ++i)
	{
		add(S , i , p1[i].c , 0);
		add(i , t1 , inf , p1[i].x + p1[i].y);
		add(i , t2 , inf , p1[i].x - p1[i].y);
		add(i , t3 , inf , -p1[i].x + p1[i].y);
		add(i , t4 , inf , -p1[i].x - p1[i].y);
	}
	
	for(int i = 1 ; i <= n ; ++i)
	{
		add(i + n , T , p2[i].c , 0);
		add(t1 , i + n , inf , -p2[i].x - p2[i].y);
		add(t2 , i + n , inf , -p2[i].x + p2[i].y);
		add(t3 , i + n , inf , p2[i].x - p2[i].y);
		add(t4 , i + n , inf , p2[i].x + p2[i].y);
	}
	
	cout << calc() << '\n'; return 0;
}
```