#include<bits/stdc++.h>
#define llong long long
#define mkpr make_pair
#define riterator reverse_iterator
using namespace std;

inline int read()
{
	int x = 0,f = 1; char ch = getchar();
	for(;!isdigit(ch);ch=getchar()) {if(ch=='-') f = -1;}
	for(; isdigit(ch);ch=getchar()) {x = x*10+ch-48;}
	return x*f;
}

const llong INF = 1e12;

namespace NetFlow
{
	const int N = 2006;
	const int M = 10000;
	struct AEdge
	{
		int u,v,wl,wr; llong c;
	} ae[M+3];
	struct Edge
	{
		int u,v,nxt,w; llong c;
	} e[(M<<1)+3];
	int fe[N+3];
	llong dis[N+3];
	int que[N+5];
	bool inq[N+3];
	int lst[N+3];
	int n,m,en,s,t; llong mf,mc;
	void addedge(int u,int v,int w,llong c)
	{
//		printf("addedge %d %d %d %lld\n",u,v,w,c);
		en++; e[en].u = u,e[en].v = v,e[en].w = w,e[en].c = c;
		e[en].nxt = fe[u]; fe[u] = en;
		en++; e[en].u = v,e[en].v = u,e[en].w = 0,e[en].c = -c;
		e[en].nxt = fe[v]; fe[v] = en;
	}
	bool spfa()
	{
		for(int i=1; i<=n; i++) dis[i] = -INF;
		int hd = 1,tl = 2; que[1] = s; dis[1] = 0;
		while(hd!=tl)
		{
			int u = que[hd]; hd++; if(hd>n+1) hd-=n+1;
			for(int i=fe[u]; i; i=e[i].nxt)
			{
				int v = e[i].v;
				if(e[i].w>0&&dis[e[i].v]<dis[u]+e[i].c)
				{
					dis[e[i].v] = dis[u]+e[i].c; lst[e[i].v] = i;
					if(!inq[e[i].v])
					{
						inq[e[i].v] = true;
						que[tl] = e[i].v; tl++; if(tl>n+1) tl-=n+1;
					}
				}
			}
			inq[u] = false;
		}
		return dis[t]!=-INF;
	}
	void calcflow()
	{
		int flow = 1e5;
		for(int u=t; u!=s; u=e[lst[u]].u)
		{
			flow = min(flow,e[lst[u]].w);
		}
		for(int u=t; u!=s; u=e[lst[u]].u)
		{
			e[lst[u]].w -= flow; e[lst[u]^1].w += flow;
		}
		mf += flow; mc += 1ll*flow*dis[t];
	}
	llong mfmc(int _n,int _s,int _t)
	{
		n = _n,s = _s,t = _t; mf = 0,mc = 0ll;
		while(spfa()) {calcflow();} return mc;
	}
}
using NetFlow::addedge;

const int N = 1000;
struct Point
{
	int x,y,cnt;
} a[N+3],b[N+3];
int n;

int main()
{
	NetFlow::en = 1;
	scanf("%d",&n);
	for(int i=1; i<=n; i++) scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].cnt);
	for(int i=1; i<=n; i++) scanf("%d%d%d",&b[i].x,&b[i].y,&b[i].cnt);
	for(int i=1; i<=n; i++)
	{
		addedge(1,i+6,a[i].cnt,0);
		addedge(i+6,3,a[i].cnt,-a[i].x-a[i].y);
		addedge(i+6,4,a[i].cnt,-a[i].x+a[i].y);
		addedge(i+6,5,a[i].cnt,a[i].x-a[i].y);
		addedge(i+6,6,a[i].cnt,a[i].x+a[i].y);
	}
	for(int i=1; i<=n; i++)
	{
		addedge(i+n+6,2,b[i].cnt,0);
		addedge(3,i+n+6,b[i].cnt,b[i].x+b[i].y);
		addedge(4,i+n+6,b[i].cnt,b[i].x-b[i].y);
		addedge(5,i+n+6,b[i].cnt,-b[i].x+b[i].y);
		addedge(6,i+n+6,b[i].cnt,-b[i].x-b[i].y);
	}
	llong ans = NetFlow::mfmc(n+n+6,1,2);
	printf("%lld\n",ans);
	return 0;
}
### Prompt

Please formulate 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<bits/stdc++.h>
#define llong long long
#define mkpr make_pair
#define riterator reverse_iterator
using namespace std;

inline int read()
{
	int x = 0,f = 1; char ch = getchar();
	for(;!isdigit(ch);ch=getchar()) {if(ch=='-') f = -1;}
	for(; isdigit(ch);ch=getchar()) {x = x*10+ch-48;}
	return x*f;
}

const llong INF = 1e12;

namespace NetFlow
{
	const int N = 2006;
	const int M = 10000;
	struct AEdge
	{
		int u,v,wl,wr; llong c;
	} ae[M+3];
	struct Edge
	{
		int u,v,nxt,w; llong c;
	} e[(M<<1)+3];
	int fe[N+3];
	llong dis[N+3];
	int que[N+5];
	bool inq[N+3];
	int lst[N+3];
	int n,m,en,s,t; llong mf,mc;
	void addedge(int u,int v,int w,llong c)
	{
//		printf("addedge %d %d %d %lld\n",u,v,w,c);
		en++; e[en].u = u,e[en].v = v,e[en].w = w,e[en].c = c;
		e[en].nxt = fe[u]; fe[u] = en;
		en++; e[en].u = v,e[en].v = u,e[en].w = 0,e[en].c = -c;
		e[en].nxt = fe[v]; fe[v] = en;
	}
	bool spfa()
	{
		for(int i=1; i<=n; i++) dis[i] = -INF;
		int hd = 1,tl = 2; que[1] = s; dis[1] = 0;
		while(hd!=tl)
		{
			int u = que[hd]; hd++; if(hd>n+1) hd-=n+1;
			for(int i=fe[u]; i; i=e[i].nxt)
			{
				int v = e[i].v;
				if(e[i].w>0&&dis[e[i].v]<dis[u]+e[i].c)
				{
					dis[e[i].v] = dis[u]+e[i].c; lst[e[i].v] = i;
					if(!inq[e[i].v])
					{
						inq[e[i].v] = true;
						que[tl] = e[i].v; tl++; if(tl>n+1) tl-=n+1;
					}
				}
			}
			inq[u] = false;
		}
		return dis[t]!=-INF;
	}
	void calcflow()
	{
		int flow = 1e5;
		for(int u=t; u!=s; u=e[lst[u]].u)
		{
			flow = min(flow,e[lst[u]].w);
		}
		for(int u=t; u!=s; u=e[lst[u]].u)
		{
			e[lst[u]].w -= flow; e[lst[u]^1].w += flow;
		}
		mf += flow; mc += 1ll*flow*dis[t];
	}
	llong mfmc(int _n,int _s,int _t)
	{
		n = _n,s = _s,t = _t; mf = 0,mc = 0ll;
		while(spfa()) {calcflow();} return mc;
	}
}
using NetFlow::addedge;

const int N = 1000;
struct Point
{
	int x,y,cnt;
} a[N+3],b[N+3];
int n;

int main()
{
	NetFlow::en = 1;
	scanf("%d",&n);
	for(int i=1; i<=n; i++) scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].cnt);
	for(int i=1; i<=n; i++) scanf("%d%d%d",&b[i].x,&b[i].y,&b[i].cnt);
	for(int i=1; i<=n; i++)
	{
		addedge(1,i+6,a[i].cnt,0);
		addedge(i+6,3,a[i].cnt,-a[i].x-a[i].y);
		addedge(i+6,4,a[i].cnt,-a[i].x+a[i].y);
		addedge(i+6,5,a[i].cnt,a[i].x-a[i].y);
		addedge(i+6,6,a[i].cnt,a[i].x+a[i].y);
	}
	for(int i=1; i<=n; i++)
	{
		addedge(i+n+6,2,b[i].cnt,0);
		addedge(3,i+n+6,b[i].cnt,b[i].x+b[i].y);
		addedge(4,i+n+6,b[i].cnt,b[i].x-b[i].y);
		addedge(5,i+n+6,b[i].cnt,-b[i].x+b[i].y);
		addedge(6,i+n+6,b[i].cnt,-b[i].x-b[i].y);
	}
	llong ans = NetFlow::mfmc(n+n+6,1,2);
	printf("%lld\n",ans);
	return 0;
}
```