#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;

const int N = 2048; typedef long long ll;

int bz[N],st[N],pre[N]; queue < int > q;
int n,tot,S,T,V1,V2,V3,V4; ll ans,dis[N];
struct EDGE{ int to,next,flow,cost; } g[N * 50];

int min(int x,int y) { return x < y ? x : y; }

void add(int u,int v,int w,int x)
{ g[++tot] = (EDGE){ v,st[u],w,x }, st[u] = tot; }

void add_edge(int u,int v,int w,int x) { add(u,v,w,x), add(v,u,0,-x); }

int spfa()
{
	memset(bz,0,sizeof bz),
	memset(pre,0,sizeof pre),
	memset(dis,-0x3f,sizeof dis);
	
	while(q.size()) q.pop();
	q.push(S), dis[S] = 0, bz[pre[S] = S] = 1;
	
	for(int x;q.size();bz[x] = 0)
	{
		x = q.front(), q.pop();
		for(int i = st[x];i;i = g[i].next)
			if(g[i].flow && dis[x] + g[i].cost > dis[g[i].to])
				!bz[g[i].to] && (q.push(g[i].to), bz[g[i].to] = 1),
				dis[g[i].to] = dis[x] + g[i].cost, pre[g[i].to] = i;
	}
	
	return pre[T];
}

void update()
{
	int flow = 0x3f3f3f3f; ll cost = 0ll;
	for(int i = T;i != S;i = g[pre[i] ^ 1].to)
		flow = min(flow,g[pre[i]].flow), cost += g[pre[i]].cost;
	ans += cost * flow;
	for(int i = T;i != S;i = g[pre[i] ^ 1].to)
		g[pre[i]].flow -= flow, g[pre[i] ^ 1].flow += flow;
}

int main()
{
	scanf("%d",&n), S = n * 2 + 1, T = S + 1, tot = 1,
	V1 = T + 1, V2 = V1 + 1, V3 = V2 + 1, V4 = V3 + 1;
	
	for(int i = 1,x,y,z;i <= n; ++ i)
		scanf("%d%d%d",&x,&y,&z), add_edge(S,i,z,0),
		add_edge(i,V1,z,x + y), add_edge(i,V2,z,x - y),
		add_edge(i,V3,z,- x + y), add_edge(i,V4,z,- x - y);
	
	for(int i = 1,x,y,z;i <= n; ++ i)
		scanf("%d%d%d",&x,&y,&z), add_edge(i + n,T,z,0),
		add_edge(V4,i + n,z,x + y), add_edge(V3,i + n,z,x - y),
		add_edge(V2,i + n,z,- x + y), add_edge(V1,i + n,z,- x - y);
	
	for(ans = 0ll;spfa();) update();
	printf("%lld\n",ans);
	
	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 <queue>
#include <cstdio>
#include <cstring>
using namespace std;

const int N = 2048; typedef long long ll;

int bz[N],st[N],pre[N]; queue < int > q;
int n,tot,S,T,V1,V2,V3,V4; ll ans,dis[N];
struct EDGE{ int to,next,flow,cost; } g[N * 50];

int min(int x,int y) { return x < y ? x : y; }

void add(int u,int v,int w,int x)
{ g[++tot] = (EDGE){ v,st[u],w,x }, st[u] = tot; }

void add_edge(int u,int v,int w,int x) { add(u,v,w,x), add(v,u,0,-x); }

int spfa()
{
	memset(bz,0,sizeof bz),
	memset(pre,0,sizeof pre),
	memset(dis,-0x3f,sizeof dis);
	
	while(q.size()) q.pop();
	q.push(S), dis[S] = 0, bz[pre[S] = S] = 1;
	
	for(int x;q.size();bz[x] = 0)
	{
		x = q.front(), q.pop();
		for(int i = st[x];i;i = g[i].next)
			if(g[i].flow && dis[x] + g[i].cost > dis[g[i].to])
				!bz[g[i].to] && (q.push(g[i].to), bz[g[i].to] = 1),
				dis[g[i].to] = dis[x] + g[i].cost, pre[g[i].to] = i;
	}
	
	return pre[T];
}

void update()
{
	int flow = 0x3f3f3f3f; ll cost = 0ll;
	for(int i = T;i != S;i = g[pre[i] ^ 1].to)
		flow = min(flow,g[pre[i]].flow), cost += g[pre[i]].cost;
	ans += cost * flow;
	for(int i = T;i != S;i = g[pre[i] ^ 1].to)
		g[pre[i]].flow -= flow, g[pre[i] ^ 1].flow += flow;
}

int main()
{
	scanf("%d",&n), S = n * 2 + 1, T = S + 1, tot = 1,
	V1 = T + 1, V2 = V1 + 1, V3 = V2 + 1, V4 = V3 + 1;
	
	for(int i = 1,x,y,z;i <= n; ++ i)
		scanf("%d%d%d",&x,&y,&z), add_edge(S,i,z,0),
		add_edge(i,V1,z,x + y), add_edge(i,V2,z,x - y),
		add_edge(i,V3,z,- x + y), add_edge(i,V4,z,- x - y);
	
	for(int i = 1,x,y,z;i <= n; ++ i)
		scanf("%d%d%d",&x,&y,&z), add_edge(i + n,T,z,0),
		add_edge(V4,i + n,z,x + y), add_edge(V3,i + n,z,x - y),
		add_edge(V2,i + n,z,- x + y), add_edge(V1,i + n,z,- x - y);
	
	for(ans = 0ll;spfa();) update();
	printf("%lld\n",ans);
	
	return 0;
}
```