#include<bits/stdc++.h>
using namespace std;
const int maxn = 20005;
typedef long long ll;
int en[maxn],nt[maxn],la[maxn],len[maxn],v[maxn],owo=1;
void adg(int x,int y,int z,int vv) {
	en[++owo]=y; nt[owo]=la[x]; la[x]=owo; len[owo]=z; v[owo]=vv;
	en[++owo]=x; nt[owo]=la[y]; la[y]=owo; len[owo]=-z; v[owo]=0;
}
int S,T,z1,z2,z3,z4;

int n;
queue<int>q;
bool rd[maxn];
int path[maxn],epath[maxn];
ll ans,dis[maxn];
bool spfa() {
	for(int i=1;i<=T;i++) dis[i] = -1e16,rd[i] = 0;
	q.push(S); dis[S] = 0;
	while(q.size()) {
		int x = q.front(); q.pop(); rd[x] = 0;
		for(int it=la[x];it;it=nt[it]) {
			int y = en[it];
			if(v[it]&&dis[y]<dis[x]+len[it]) {
				dis[y] = dis[x] + len[it];
				path[y] = x;
				epath[y] = it;
				if(!rd[y]) q.push(y),rd[y] = 1;
			}
		}
	}
	if(dis[T]==(ll)-1e16) return 0;
	int mi = 0x3f3f3f3f;
	for(int p=T;p!=S;p=path[p]) mi = min(mi,v[epath[p]]);
	ans += 1ll * mi * dis[T];
	for(int p=T;p!=S;p=path[p]) {
		v[epath[p]] -= mi;
		v[epath[p]^1] += mi;
	}
	return 1;
}

int main() {
	scanf("%d",&n);
	z1 = n+n+1; z2 = n+n+2; z3 = n+n+3; z4 = n+n+4;
	S = n+n+5; T = n+n+6;
	for(int i=1;i<=n;i++) {
		int x,y,c; scanf("%d%d%d",&x,&y,&c);
		adg(i,z1,x+y,0x3f3f3f3f);
		adg(i,z2,-x+y,0x3f3f3f3f);
		adg(i,z3,x-y,0x3f3f3f3f);
		adg(i,z4,-x-y,0x3f3f3f3f);
		adg(S,i,0,c);
	}
	for(int i=1;i<=n;i++) {
		int x,y,c; scanf("%d%d%d",&x,&y,&c);
		adg(z1,i+n,-x-y,0x3f3f3f3f);
		adg(z2,i+n,x-y,0x3f3f3f3f);
		adg(z3,i+n,-x+y,0x3f3f3f3f);
		adg(z4,i+n,x+y,0x3f3f3f3f);
		adg(i+n,T,0,c);
	}
	while(spfa());
	printf("%lld",ans);
}
### Prompt

Develop a solution in Cpp to the problem described below:

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>
using namespace std;
const int maxn = 20005;
typedef long long ll;
int en[maxn],nt[maxn],la[maxn],len[maxn],v[maxn],owo=1;
void adg(int x,int y,int z,int vv) {
	en[++owo]=y; nt[owo]=la[x]; la[x]=owo; len[owo]=z; v[owo]=vv;
	en[++owo]=x; nt[owo]=la[y]; la[y]=owo; len[owo]=-z; v[owo]=0;
}
int S,T,z1,z2,z3,z4;

int n;
queue<int>q;
bool rd[maxn];
int path[maxn],epath[maxn];
ll ans,dis[maxn];
bool spfa() {
	for(int i=1;i<=T;i++) dis[i] = -1e16,rd[i] = 0;
	q.push(S); dis[S] = 0;
	while(q.size()) {
		int x = q.front(); q.pop(); rd[x] = 0;
		for(int it=la[x];it;it=nt[it]) {
			int y = en[it];
			if(v[it]&&dis[y]<dis[x]+len[it]) {
				dis[y] = dis[x] + len[it];
				path[y] = x;
				epath[y] = it;
				if(!rd[y]) q.push(y),rd[y] = 1;
			}
		}
	}
	if(dis[T]==(ll)-1e16) return 0;
	int mi = 0x3f3f3f3f;
	for(int p=T;p!=S;p=path[p]) mi = min(mi,v[epath[p]]);
	ans += 1ll * mi * dis[T];
	for(int p=T;p!=S;p=path[p]) {
		v[epath[p]] -= mi;
		v[epath[p]^1] += mi;
	}
	return 1;
}

int main() {
	scanf("%d",&n);
	z1 = n+n+1; z2 = n+n+2; z3 = n+n+3; z4 = n+n+4;
	S = n+n+5; T = n+n+6;
	for(int i=1;i<=n;i++) {
		int x,y,c; scanf("%d%d%d",&x,&y,&c);
		adg(i,z1,x+y,0x3f3f3f3f);
		adg(i,z2,-x+y,0x3f3f3f3f);
		adg(i,z3,x-y,0x3f3f3f3f);
		adg(i,z4,-x-y,0x3f3f3f3f);
		adg(S,i,0,c);
	}
	for(int i=1;i<=n;i++) {
		int x,y,c; scanf("%d%d%d",&x,&y,&c);
		adg(z1,i+n,-x-y,0x3f3f3f3f);
		adg(z2,i+n,x-y,0x3f3f3f3f);
		adg(z3,i+n,-x+y,0x3f3f3f3f);
		adg(z4,i+n,x+y,0x3f3f3f3f);
		adg(i+n,T,0,c);
	}
	while(spfa());
	printf("%lld",ans);
}
```