#include <cstdio>
#include <cassert>
#include <cstring>
#include <algorithm>
const int N=2006, M=20000;
int n, m, head[N], to[M], cap[M], cost[M], next[M], queue[65536], prev[M];
bool vis[N];
long long dis[N];
template<class T1, class T2> bool cmin(T1 &a, const T2 &b) {return b<a?(a=b, true):false;}
template<class T1, class T2> bool cmax(T1 &a, const T2 &b) {return a<b?(a=b, true):false;}
void add_edge(int u, int v, int mx, int w) {
	to[m]=v, cap[m]=mx, cost[m]=w, next[m]=head[u], head[u]=m++;
	to[m]=u, cap[m]=0, cost[m]=-w, next[m]=head[v], head[v]=m++;
}
int main() {
	assert(scanf("%d", &n)==1);
	memset(head, -1, (2*n+6)*sizeof(int));
	for(int i=6; i<n+6; ++i) {
		int x, y, c;
		assert(scanf("%d%d%d", &x, &y, &c)==3);
		add_edge(0, i, c, 0);
		add_edge(i, 2, c, x+y);
		add_edge(i, 3, c, x-y);
		add_edge(i, 4, c,-x+y);
		add_edge(i, 5, c,-x-y);
	}
	for(int i=n+6; i<2*n+6; ++i) {
		int x, y, c;
		assert(scanf("%d%d%d", &x, &y, &c)==3);
		add_edge(i, 1, c, 0);
		add_edge(5, i, c, x+y);
		add_edge(4, i, c, x-y);
		add_edge(3, i, c,-x+y);
		add_edge(2, i, c,-x-y);
	}
	long long ans=0;
	while(true) {
		unsigned short ql=0, qr=0;
		memset(dis+1, 192, (2*n+5)*sizeof(long long));
		queue[qr++]=0;
		while(ql!=qr) {
			int v=queue[ql++];
			vis[v]=false;
			for(int i=head[v]; ~i; i=next[i])
				if(cap[i]&&cmax(dis[to[i]], dis[v]+cost[i])) {
					prev[to[i]]=i;
					if(!vis[to[i]]) vis[queue[qr++]=to[i]]=true;
				}
		}
		if(dis[1]<0) break;
		int flow=1e9;
		for(int i=1; i!=0; i=to[prev[i]^1]) cmin(flow, cap[prev[i]]);
		for(int i=1; i!=0; i=to[prev[i]^1]) cap[prev[i]]-=flow, cap[prev[i]^1]+=flow;
		ans+=dis[1]*flow;
	}
	printf("%lld\n", ans);
	return 0;
}

### Prompt

Your task is to create 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 <cassert>
#include <cstring>
#include <algorithm>
const int N=2006, M=20000;
int n, m, head[N], to[M], cap[M], cost[M], next[M], queue[65536], prev[M];
bool vis[N];
long long dis[N];
template<class T1, class T2> bool cmin(T1 &a, const T2 &b) {return b<a?(a=b, true):false;}
template<class T1, class T2> bool cmax(T1 &a, const T2 &b) {return a<b?(a=b, true):false;}
void add_edge(int u, int v, int mx, int w) {
	to[m]=v, cap[m]=mx, cost[m]=w, next[m]=head[u], head[u]=m++;
	to[m]=u, cap[m]=0, cost[m]=-w, next[m]=head[v], head[v]=m++;
}
int main() {
	assert(scanf("%d", &n)==1);
	memset(head, -1, (2*n+6)*sizeof(int));
	for(int i=6; i<n+6; ++i) {
		int x, y, c;
		assert(scanf("%d%d%d", &x, &y, &c)==3);
		add_edge(0, i, c, 0);
		add_edge(i, 2, c, x+y);
		add_edge(i, 3, c, x-y);
		add_edge(i, 4, c,-x+y);
		add_edge(i, 5, c,-x-y);
	}
	for(int i=n+6; i<2*n+6; ++i) {
		int x, y, c;
		assert(scanf("%d%d%d", &x, &y, &c)==3);
		add_edge(i, 1, c, 0);
		add_edge(5, i, c, x+y);
		add_edge(4, i, c, x-y);
		add_edge(3, i, c,-x+y);
		add_edge(2, i, c,-x-y);
	}
	long long ans=0;
	while(true) {
		unsigned short ql=0, qr=0;
		memset(dis+1, 192, (2*n+5)*sizeof(long long));
		queue[qr++]=0;
		while(ql!=qr) {
			int v=queue[ql++];
			vis[v]=false;
			for(int i=head[v]; ~i; i=next[i])
				if(cap[i]&&cmax(dis[to[i]], dis[v]+cost[i])) {
					prev[to[i]]=i;
					if(!vis[to[i]]) vis[queue[qr++]=to[i]]=true;
				}
		}
		if(dis[1]<0) break;
		int flow=1e9;
		for(int i=1; i!=0; i=to[prev[i]^1]) cmin(flow, cap[prev[i]]);
		for(int i=1; i!=0; i=to[prev[i]^1]) cap[prev[i]]-=flow, cap[prev[i]^1]+=flow;
		ans+=dis[1]*flow;
	}
	printf("%lld\n", ans);
	return 0;
}

```