#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
const ll N=1e5+10;
ll w[N],ne[N],la[N],len[N],cst[N],t,n;
ll s[N],cur[N];
ll d[N],dep[N],ti,vis[N],S,T,d1[N],d2[N],p1,p2,p3,p4;
void alink(ll x,ll y,ll z,ll l){
	w[++t]=y;
	ne[t]=la[x];
	la[x]=t;
	len[t]=z;
	cst[t]=l;
}
void link(ll x,ll y,ll z,ll l){
	alink(x,y,z,l);
	alink(y,x,0,-l);
}
ll spfa(){
	for (ll i=1;i<=T;i++)s[i]=-1e18,vis[i]=0;
	s[S]=0;vis[S]=1;d[1]=S;
	ll l=0,r=1;
	while (l<r){
		ll x=d[++l];
		for (ll y=la[x];y;y=ne[y]){
			ll z=w[y];
			if (len[y]&&s[z]<s[x]+cst[y]){
				s[z]=s[x]+cst[y];
				if (!vis[z]){
					vis[z]=1;
					d[++r]=z;
				}
			}
		}
		vis[x]=0;
	}
	vis[S]=1;l=0,r=1;
	while (l<r){
		ll x=d[++l];
		cur[x]=la[x];
		for (ll y=la[x];y;y=ne[y]){
			ll z=w[y];
			if (len[y]&&s[z]==s[x]+cst[y]&&!vis[z]){
				vis[z]=1;
				d[++r]=z;
				dep[z]=dep[x]+1;
			}
		}
	}
	return vis[T];
}
ll ans;
ll flow(ll x,ll val){
	if (x==T)return val;
	ll v1=val;
	for (ll y=cur[x];y;y=ne[y]){
		cur[x]=y;
		ll z=w[y];
		if (dep[z]!=dep[x]+1||!len[y]||s[z]!=s[x]+cst[y])continue;
		ll use=flow(z,min(val,len[y]));
		ans+=(ll)cst[y]*use;
		val-=use;
		len[y]-=use;
		len[y^1]+=use;
		if (!val)return v1;
	}return v1-val;
}
int main(){
	cin>>n;
	for (ll i=1;i<=n;i++)d1[i]=++t,d2[i]=++t;
	p1=++t,p2=++t,p3=++t,p4=++t;
	S=++t;T=++t;t=1;
	for (ll i=1;i<=n;i++){
		ll x,y,z;
		scanf("%lld%lld%lld",&x,&y,&z);
		link(S,d1[i],z,0);
		link(d1[i],p1,z,x+y);
		link(d1[i],p2,z,x-y);
		link(d1[i],p3,z,y-x);
		link(d1[i],p4,z,-x-y);
	}
	for (ll i=1;i<=n;i++){
		ll x,y,z;
		scanf("%lld%lld%lld",&x,&y,&z);
		link(d2[i],T,z,0);
		link(p1,d2[i],z,-x-y);
		link(p2,d2[i],z,y-x);
		link(p3,d2[i],z,x-y);
		link(p4,d2[i],z,x+y);
	}
	while (spfa())flow(S,1<<30);
	cout<<ans<<endl;
}
### 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<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
const ll N=1e5+10;
ll w[N],ne[N],la[N],len[N],cst[N],t,n;
ll s[N],cur[N];
ll d[N],dep[N],ti,vis[N],S,T,d1[N],d2[N],p1,p2,p3,p4;
void alink(ll x,ll y,ll z,ll l){
	w[++t]=y;
	ne[t]=la[x];
	la[x]=t;
	len[t]=z;
	cst[t]=l;
}
void link(ll x,ll y,ll z,ll l){
	alink(x,y,z,l);
	alink(y,x,0,-l);
}
ll spfa(){
	for (ll i=1;i<=T;i++)s[i]=-1e18,vis[i]=0;
	s[S]=0;vis[S]=1;d[1]=S;
	ll l=0,r=1;
	while (l<r){
		ll x=d[++l];
		for (ll y=la[x];y;y=ne[y]){
			ll z=w[y];
			if (len[y]&&s[z]<s[x]+cst[y]){
				s[z]=s[x]+cst[y];
				if (!vis[z]){
					vis[z]=1;
					d[++r]=z;
				}
			}
		}
		vis[x]=0;
	}
	vis[S]=1;l=0,r=1;
	while (l<r){
		ll x=d[++l];
		cur[x]=la[x];
		for (ll y=la[x];y;y=ne[y]){
			ll z=w[y];
			if (len[y]&&s[z]==s[x]+cst[y]&&!vis[z]){
				vis[z]=1;
				d[++r]=z;
				dep[z]=dep[x]+1;
			}
		}
	}
	return vis[T];
}
ll ans;
ll flow(ll x,ll val){
	if (x==T)return val;
	ll v1=val;
	for (ll y=cur[x];y;y=ne[y]){
		cur[x]=y;
		ll z=w[y];
		if (dep[z]!=dep[x]+1||!len[y]||s[z]!=s[x]+cst[y])continue;
		ll use=flow(z,min(val,len[y]));
		ans+=(ll)cst[y]*use;
		val-=use;
		len[y]-=use;
		len[y^1]+=use;
		if (!val)return v1;
	}return v1-val;
}
int main(){
	cin>>n;
	for (ll i=1;i<=n;i++)d1[i]=++t,d2[i]=++t;
	p1=++t,p2=++t,p3=++t,p4=++t;
	S=++t;T=++t;t=1;
	for (ll i=1;i<=n;i++){
		ll x,y,z;
		scanf("%lld%lld%lld",&x,&y,&z);
		link(S,d1[i],z,0);
		link(d1[i],p1,z,x+y);
		link(d1[i],p2,z,x-y);
		link(d1[i],p3,z,y-x);
		link(d1[i],p4,z,-x-y);
	}
	for (ll i=1;i<=n;i++){
		ll x,y,z;
		scanf("%lld%lld%lld",&x,&y,&z);
		link(d2[i],T,z,0);
		link(p1,d2[i],z,-x-y);
		link(p2,d2[i],z,y-x);
		link(p3,d2[i],z,x-y);
		link(p4,d2[i],z,x+y);
	}
	while (spfa())flow(S,1<<30);
	cout<<ans<<endl;
}
```