#include <stdio.h>
#include <algorithm>
#include <queue>

using namespace std;

//long long
struct node {
	long long v , f , c;
	node *next , *rev;
} pool[1200000*3] , *g[12000] , *from[12000];
long long top;
long long n;
long long rx[1200] , ry[1200] , rc[1200];
long long bx[1200] , by[1200] , bc[1200];
long long dis[12000] , vis[12000];
long long s , t;
void add ( long long u , long long v , long long f , long long c ) {
	node *tmp1 = &pool[++top] , *tmp2 = &pool[++top];
	//9printf ( "%lld %lld %lld %lld\n" , u , v , f , c );
	tmp1 -> v = v; tmp1 -> f = f; tmp1 -> c = c; tmp1 -> next = g[u]; g[u] = tmp1; tmp1 -> rev = tmp2;
	tmp2 -> v = u; tmp2 -> f = 0; tmp2 -> c = -c; tmp2 -> next = g[v]; g[v] = tmp2; tmp2 -> rev = tmp1;
}
bool spfa () {
	long long i , k;
	queue < long long > q;
	for ( i = 1 ; i <= t ; i++ ) {
		dis[i] = -2147483647777777ll;
		vis[i] = 0;
	}
	dis[s] = 0; vis[s] = 1;
	q.push ( s );
	while ( q.size () ) {
		k = q.front (); q.pop ();
		vis[k] = 0;
		for ( node *j = g[k] ; j ; j = j -> next ) {
			//printf ( "%lld %lld\n" , j -> f , j -> c );
			if ( j -> f && dis[j->v] < dis[k] + j -> c ) {
				dis[j->v] = dis[k] + j -> c;
				from[j->v] = j;
				if ( !vis[j->v] ) q.push ( j -> v );
				vis[j->v] = 1;
			}
		}
	}
	if ( dis[t] != -2147483647777777ll ) return true;
	return false;
}
long long find () {
	long long i , minn = 2147483647777777ll;
	for ( i = t ; i != s ; i = from[i] -> rev -> v ) minn = min ( minn , from[i] -> f );
	for ( i = t ; i != s ; i = from[i] -> rev -> v ) {
		from[i] -> f -= minn;
		from[i] -> rev -> f += minn;
	}
	return dis[t] * minn;
}
void dinic () {
	long long ans = 0;
	while ( spfa () == true ) ans += find ();
	printf ( "%lld\n" , ans );
}
void work () {
	int i;
	scanf ( "%lld" , &n );
	for ( i = 1 ; i <= n ; i++ ) {
		scanf ( "%lld%lld%lld" , &rx[i] , &ry[i] , &rc[i] );
	}
	for ( i = 1 ; i <= n ; i++ ) {
		scanf ( "%lld%lld%lld" , &bx[i] , &by[i] , &bc[i] );
	}
	for ( i = 1 ; i <= n ; i++ ) {
		add ( 1 , 1 + i , rc[i] , 0 );
		add ( 1 + n + 4 + i , 1 + n + 4 + n + 1 , bc[i] , 0 );
		add ( 1 + i , 1 + n + 1 , 10 , - rx[i] - ry[i] );
		add ( 1 + i , 1 + n + 2 , 10 , - rx[i] + ry[i] );
		add ( 1 + i , 1 + n + 3 , 10 , + rx[i] - ry[i] );
		add ( 1 + i , 1 + n + 4 , 10 , + rx[i] + ry[i] );
		add ( 1 + n + 1 , 1 + n + 4 + i , 10 , + bx[i] + by[i] );
		add ( 1 + n + 2 , 1 + n + 4 + i , 10 , + bx[i] - by[i] );
		add ( 1 + n + 3 , 1 + n + 4 + i , 10 , - bx[i] + by[i] );
		add ( 1 + n + 4 , 1 + n + 4 + i , 10 , - bx[i] - by[i] );
	}
	s = 1; t = 1 + n + 4 + n + 1;
	dinic ();
}
int main () {
	work ();
	return 0;
}

### Prompt

In Cpp, your task is to solve 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 <stdio.h>
#include <algorithm>
#include <queue>

using namespace std;

//long long
struct node {
	long long v , f , c;
	node *next , *rev;
} pool[1200000*3] , *g[12000] , *from[12000];
long long top;
long long n;
long long rx[1200] , ry[1200] , rc[1200];
long long bx[1200] , by[1200] , bc[1200];
long long dis[12000] , vis[12000];
long long s , t;
void add ( long long u , long long v , long long f , long long c ) {
	node *tmp1 = &pool[++top] , *tmp2 = &pool[++top];
	//9printf ( "%lld %lld %lld %lld\n" , u , v , f , c );
	tmp1 -> v = v; tmp1 -> f = f; tmp1 -> c = c; tmp1 -> next = g[u]; g[u] = tmp1; tmp1 -> rev = tmp2;
	tmp2 -> v = u; tmp2 -> f = 0; tmp2 -> c = -c; tmp2 -> next = g[v]; g[v] = tmp2; tmp2 -> rev = tmp1;
}
bool spfa () {
	long long i , k;
	queue < long long > q;
	for ( i = 1 ; i <= t ; i++ ) {
		dis[i] = -2147483647777777ll;
		vis[i] = 0;
	}
	dis[s] = 0; vis[s] = 1;
	q.push ( s );
	while ( q.size () ) {
		k = q.front (); q.pop ();
		vis[k] = 0;
		for ( node *j = g[k] ; j ; j = j -> next ) {
			//printf ( "%lld %lld\n" , j -> f , j -> c );
			if ( j -> f && dis[j->v] < dis[k] + j -> c ) {
				dis[j->v] = dis[k] + j -> c;
				from[j->v] = j;
				if ( !vis[j->v] ) q.push ( j -> v );
				vis[j->v] = 1;
			}
		}
	}
	if ( dis[t] != -2147483647777777ll ) return true;
	return false;
}
long long find () {
	long long i , minn = 2147483647777777ll;
	for ( i = t ; i != s ; i = from[i] -> rev -> v ) minn = min ( minn , from[i] -> f );
	for ( i = t ; i != s ; i = from[i] -> rev -> v ) {
		from[i] -> f -= minn;
		from[i] -> rev -> f += minn;
	}
	return dis[t] * minn;
}
void dinic () {
	long long ans = 0;
	while ( spfa () == true ) ans += find ();
	printf ( "%lld\n" , ans );
}
void work () {
	int i;
	scanf ( "%lld" , &n );
	for ( i = 1 ; i <= n ; i++ ) {
		scanf ( "%lld%lld%lld" , &rx[i] , &ry[i] , &rc[i] );
	}
	for ( i = 1 ; i <= n ; i++ ) {
		scanf ( "%lld%lld%lld" , &bx[i] , &by[i] , &bc[i] );
	}
	for ( i = 1 ; i <= n ; i++ ) {
		add ( 1 , 1 + i , rc[i] , 0 );
		add ( 1 + n + 4 + i , 1 + n + 4 + n + 1 , bc[i] , 0 );
		add ( 1 + i , 1 + n + 1 , 10 , - rx[i] - ry[i] );
		add ( 1 + i , 1 + n + 2 , 10 , - rx[i] + ry[i] );
		add ( 1 + i , 1 + n + 3 , 10 , + rx[i] - ry[i] );
		add ( 1 + i , 1 + n + 4 , 10 , + rx[i] + ry[i] );
		add ( 1 + n + 1 , 1 + n + 4 + i , 10 , + bx[i] + by[i] );
		add ( 1 + n + 2 , 1 + n + 4 + i , 10 , + bx[i] - by[i] );
		add ( 1 + n + 3 , 1 + n + 4 + i , 10 , - bx[i] + by[i] );
		add ( 1 + n + 4 , 1 + n + 4 + i , 10 , - bx[i] - by[i] );
	}
	s = 1; t = 1 + n + 4 + n + 1;
	dinic ();
}
int main () {
	work ();
	return 0;
}

```