#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn = 1000 + 10;
const int N = 200000 + 10;
const int E = 1000000 + 10;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int DEBUG = 0;

ll flow, cost;
int S, T, n, h[N], ecnt;
struct enode{
	int v, n;
	ll w, c;
	int op;
	enode() {}
	enode(int _v, int _n, ll _w, ll _c, int _op):v(_v), n(_n), w(_w), c(_c), op(_op) {}
}e[E << 1];

inline void addedge(int u, int v, ll w, ll c) {
	c = - c;
	if(DEBUG) printf("%d %d %lld %lld\n", u, v, w, c);
	ecnt ++; e[ecnt] = enode(v,h[u],w,c,ecnt + 1); h[u] = ecnt;
	ecnt ++; e[ecnt] = enode(u,h[v],0,-c,ecnt - 1); h[v] = ecnt;
}

ll d[N];
int pre[N];
bool inq[N];
queue<int> q;

inline bool spfa() {
	for(int i = S;i <= T;i ++) d[i] = - INF;
	memset(inq,false,sizeof(inq));
	memset(pre,-1,sizeof(pre));
	while(!q.empty()) q.pop();
	q.push(S); d[S] = 0; inq[S] = 1; pre[S] = -1;
	while(!q.empty()) {
		int u = q.front(); q.pop();
		for(int i = h[u];~ i;i = e[i].n) {
			int v = e[i].v;
			if(e[i].w > 0 && d[v] < d[u] + e[i].c) {
				d[v] = d[u] + e[i].c;
				pre[v] = i;
				if(!inq[v]) {
					inq[v] = 1;
					q.push(v);
				}
			}
		}
		inq[u] = 0;
	}
	if(d[T] == - INF) return 0;
	ll tmp = INF, now = 0;
	for(int i = pre[T];~ i;i = pre[e[e[i].op].v]) {
		tmp = min(tmp,(ll)e[i].w);
	}
	for(int i = pre[T];~ i;i = pre[e[e[i].op].v]) {
		e[i].w -= tmp;
		e[e[i].op].w += tmp;
		now += (ll)tmp * e[i].c;
	}
	flow += tmp; cost += now;
	return 1;
}

inline void mcmf() {
	flow = cost = 0;
	while(spfa());
}

struct anode{
	int x, y, w;
	anode() {}
	anode(int _x, int _y, int _w):x(_x), y(_y), w(_w) {}
}a[maxn], b[maxn];
int Tmp[5];

int main() {
	scanf("%d", &n); ecnt = 0;
	memset(h,-1,sizeof(h)); 
	for(int i = 1;i <= n;i ++) {
		int x, y, w;
		scanf("%d%d%d", &x, &y, &w);
		a[i] = anode(x,y,w);
	}
	for(int i = 1;i <= n;i ++) {
		int x, y, w;
		scanf("%d%d%d", &x, &y, &w);
		b[i] = anode(x,y,w);
	}
	S = 0, T = n + n + 4 + 1;
	for(int i = 1;i <= 4;i ++) Tmp[i] = n + n + i;
	for(int i = 1;i <= n;i ++) {
		addedge(S,i,a[i].w,0);
		addedge(i,Tmp[1],INF,(ll)+ a[i].x + a[i].y);
		addedge(i,Tmp[2],INF,(ll)+ a[i].x - a[i].y);
		addedge(i,Tmp[3],INF,(ll)- a[i].x + a[i].y);
		addedge(i,Tmp[4],INF,(ll)- a[i].x - a[i].y);
	}
	for(int i = 1;i <= n;i ++) {
		addedge(i + n,T,b[i].w,0);
		addedge(Tmp[1],i + n,INF,(ll)- b[i].x - b[i].y);
		addedge(Tmp[2],i + n,INF,(ll)- b[i].x + b[i].y);
		addedge(Tmp[3],i + n,INF,(ll)+ b[i].x - b[i].y);
		addedge(Tmp[4],i + n,INF,(ll)+ b[i].x + b[i].y);
	}
	mcmf();
	printf("%lld\n", cost);
	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<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn = 1000 + 10;
const int N = 200000 + 10;
const int E = 1000000 + 10;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int DEBUG = 0;

ll flow, cost;
int S, T, n, h[N], ecnt;
struct enode{
	int v, n;
	ll w, c;
	int op;
	enode() {}
	enode(int _v, int _n, ll _w, ll _c, int _op):v(_v), n(_n), w(_w), c(_c), op(_op) {}
}e[E << 1];

inline void addedge(int u, int v, ll w, ll c) {
	c = - c;
	if(DEBUG) printf("%d %d %lld %lld\n", u, v, w, c);
	ecnt ++; e[ecnt] = enode(v,h[u],w,c,ecnt + 1); h[u] = ecnt;
	ecnt ++; e[ecnt] = enode(u,h[v],0,-c,ecnt - 1); h[v] = ecnt;
}

ll d[N];
int pre[N];
bool inq[N];
queue<int> q;

inline bool spfa() {
	for(int i = S;i <= T;i ++) d[i] = - INF;
	memset(inq,false,sizeof(inq));
	memset(pre,-1,sizeof(pre));
	while(!q.empty()) q.pop();
	q.push(S); d[S] = 0; inq[S] = 1; pre[S] = -1;
	while(!q.empty()) {
		int u = q.front(); q.pop();
		for(int i = h[u];~ i;i = e[i].n) {
			int v = e[i].v;
			if(e[i].w > 0 && d[v] < d[u] + e[i].c) {
				d[v] = d[u] + e[i].c;
				pre[v] = i;
				if(!inq[v]) {
					inq[v] = 1;
					q.push(v);
				}
			}
		}
		inq[u] = 0;
	}
	if(d[T] == - INF) return 0;
	ll tmp = INF, now = 0;
	for(int i = pre[T];~ i;i = pre[e[e[i].op].v]) {
		tmp = min(tmp,(ll)e[i].w);
	}
	for(int i = pre[T];~ i;i = pre[e[e[i].op].v]) {
		e[i].w -= tmp;
		e[e[i].op].w += tmp;
		now += (ll)tmp * e[i].c;
	}
	flow += tmp; cost += now;
	return 1;
}

inline void mcmf() {
	flow = cost = 0;
	while(spfa());
}

struct anode{
	int x, y, w;
	anode() {}
	anode(int _x, int _y, int _w):x(_x), y(_y), w(_w) {}
}a[maxn], b[maxn];
int Tmp[5];

int main() {
	scanf("%d", &n); ecnt = 0;
	memset(h,-1,sizeof(h)); 
	for(int i = 1;i <= n;i ++) {
		int x, y, w;
		scanf("%d%d%d", &x, &y, &w);
		a[i] = anode(x,y,w);
	}
	for(int i = 1;i <= n;i ++) {
		int x, y, w;
		scanf("%d%d%d", &x, &y, &w);
		b[i] = anode(x,y,w);
	}
	S = 0, T = n + n + 4 + 1;
	for(int i = 1;i <= 4;i ++) Tmp[i] = n + n + i;
	for(int i = 1;i <= n;i ++) {
		addedge(S,i,a[i].w,0);
		addedge(i,Tmp[1],INF,(ll)+ a[i].x + a[i].y);
		addedge(i,Tmp[2],INF,(ll)+ a[i].x - a[i].y);
		addedge(i,Tmp[3],INF,(ll)- a[i].x + a[i].y);
		addedge(i,Tmp[4],INF,(ll)- a[i].x - a[i].y);
	}
	for(int i = 1;i <= n;i ++) {
		addedge(i + n,T,b[i].w,0);
		addedge(Tmp[1],i + n,INF,(ll)- b[i].x - b[i].y);
		addedge(Tmp[2],i + n,INF,(ll)- b[i].x + b[i].y);
		addedge(Tmp[3],i + n,INF,(ll)+ b[i].x - b[i].y);
		addedge(Tmp[4],i + n,INF,(ll)+ b[i].x + b[i].y);
	}
	mcmf();
	printf("%lld\n", cost);
	return 0;
}
```