#include <cstdio>
#include <cstring>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
#define rep(i,a) for(int i=lst[a];i;i=nxt[i])
using namespace std;

typedef long long ll;

const int N=2e3+10,M=2e4+10,inf=0x7fffffff;

int t[M],f[M],v[M],nxt[M],lst[N],l;

void add(int x,int y,int z,int c) {
	t[++l]=y;f[l]=z;v[l]=c;nxt[l]=lst[x];lst[x]=l;
	t[++l]=x;f[l]=0;v[l]=-c;nxt[l]=lst[y];lst[y]=l;
}

int n,S,T,t1,t2,t3,t4,x,y,c,q[M],pre[N];
ll dis[N],ans=0;
bool vis[N];

bool spfa() {
	fo(i,S,t4) dis[i]=-1e12;dis[S]=0;
	int i=0,j=1;q[1]=S;vis[S]=1;
	while (i<j) {
		rep(k,q[++i])
			if (f[k]&&dis[t[k]]<dis[q[i]]+v[k]) {
				dis[t[k]]=dis[q[i]]+v[k];pre[t[k]]=k;
				if (!vis[t[k]]) vis[t[k]]=1,q[++j]=t[k];
			}
		vis[q[i]]=0;
	}
	return dis[T]>-1e12;
}

void find() {
	int flow=inf;
	for(int x=T;x!=S;x=t[pre[x]^1]) flow=min(flow,f[pre[x]]);
	for(int x=T;x!=S;x=t[pre[x]^1]) {
		f[pre[x]]-=flow;f[pre[x]^1]+=flow;
		ans+=(ll)flow*v[pre[x]];
	}
}

int main() {
	scanf("%d",&n);l=1;
	S=0;T=(n<<1)+1;t1=T+1;t2=t1+1;t3=t2+1;t4=t3+1;
	fo(i,1,n) {
		scanf("%d%d%d",&x,&y,&c);
		add(S,i,c,0);
		add(i,t1,inf,x+y);
		add(i,t2,inf,x-y);
		add(i,t3,inf,-x+y);
		add(i,t4,inf,-x-y);
	}
	fo(i,1,n) {
		scanf("%d%d%d",&x,&y,&c);
		add(i+n,T,c,0);
		add(t1,i+n,inf,-x-y);
		add(t2,i+n,inf,-x+y);
		add(t3,i+n,inf,x-y);
		add(t4,i+n,inf,x+y);
	}
	while (spfa()) find();
	printf("%lld\n",ans);
	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 <cstdio>
#include <cstring>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
#define rep(i,a) for(int i=lst[a];i;i=nxt[i])
using namespace std;

typedef long long ll;

const int N=2e3+10,M=2e4+10,inf=0x7fffffff;

int t[M],f[M],v[M],nxt[M],lst[N],l;

void add(int x,int y,int z,int c) {
	t[++l]=y;f[l]=z;v[l]=c;nxt[l]=lst[x];lst[x]=l;
	t[++l]=x;f[l]=0;v[l]=-c;nxt[l]=lst[y];lst[y]=l;
}

int n,S,T,t1,t2,t3,t4,x,y,c,q[M],pre[N];
ll dis[N],ans=0;
bool vis[N];

bool spfa() {
	fo(i,S,t4) dis[i]=-1e12;dis[S]=0;
	int i=0,j=1;q[1]=S;vis[S]=1;
	while (i<j) {
		rep(k,q[++i])
			if (f[k]&&dis[t[k]]<dis[q[i]]+v[k]) {
				dis[t[k]]=dis[q[i]]+v[k];pre[t[k]]=k;
				if (!vis[t[k]]) vis[t[k]]=1,q[++j]=t[k];
			}
		vis[q[i]]=0;
	}
	return dis[T]>-1e12;
}

void find() {
	int flow=inf;
	for(int x=T;x!=S;x=t[pre[x]^1]) flow=min(flow,f[pre[x]]);
	for(int x=T;x!=S;x=t[pre[x]^1]) {
		f[pre[x]]-=flow;f[pre[x]^1]+=flow;
		ans+=(ll)flow*v[pre[x]];
	}
}

int main() {
	scanf("%d",&n);l=1;
	S=0;T=(n<<1)+1;t1=T+1;t2=t1+1;t3=t2+1;t4=t3+1;
	fo(i,1,n) {
		scanf("%d%d%d",&x,&y,&c);
		add(S,i,c,0);
		add(i,t1,inf,x+y);
		add(i,t2,inf,x-y);
		add(i,t3,inf,-x+y);
		add(i,t4,inf,-x-y);
	}
	fo(i,1,n) {
		scanf("%d%d%d",&x,&y,&c);
		add(i+n,T,c,0);
		add(t1,i+n,inf,-x-y);
		add(t2,i+n,inf,-x+y);
		add(t3,i+n,inf,x-y);
		add(t4,i+n,inf,x+y);
	}
	while (spfa()) find();
	printf("%lld\n",ans);
	return 0;
}
```