#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 2005
#define ba 47
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 +c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
int S,T,Ncnt,N;
int a[2][MAXN],ty[4];
struct node {
    int to,next,cap;int64 val;
}E[MAXN * 100];
int head[MAXN * 2],sumE = 1;
int rx[MAXN],ry[MAXN],rc[MAXN];
int bx[MAXN],by[MAXN],bc[MAXN];
int dx[4] = {1,1,-1,-1};
int dy[4] = {1,-1,1,-1};

void add(int u,int v,int c,int64 a) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    E[sumE].cap = c;
    E[sumE].val = a;
    head[u] = sumE;
}
void addtwo(int u,int v,int c,int64 a) {
    add(u,v,c,a);
    add(v,u,0,-a);
}
int64 dis[MAXN];bool inq[MAXN];
int preE[MAXN];
queue<int> Q;
int64 SPFA() {
    for(int i = 1 ; i <= Ncnt ; ++i) dis[i] = -1e18;
    memset(inq,0,sizeof(inq));
    dis[S] = 0;inq[S] = 1;Q.push(S);
    while(!Q.empty()) {
	int u = Q.front();Q.pop();inq[u] = 0;
	for(int i = head[u] ; i ; i = E[i].next) {
	    int v = E[i].to;
	    if(E[i].cap) {
		if(dis[v] < dis[u] + E[i].val) {
		    dis[v] = dis[u] + E[i].val;
		    preE[v] = i;
		    if(!inq[v]) {Q.push(v);inq[v] = 1;}
		}
	    }
	}
    }
    return dis[T];
}
void Init() {
    read(N);
    S = ++Ncnt;
    for(int i = 0 ; i < 2 ; ++i) {
	for(int j = 1 ; j <= N ; ++j) {
	    a[i][j] = ++Ncnt;
	}
    }
    for(int i = 0 ; i < 4 ; ++i) ty[i] = ++Ncnt;
    T = ++Ncnt;
    for(int i = 1 ; i <= N ; ++i) {read(rx[i]);read(ry[i]);read(rc[i]);}
    for(int i = 1 ; i <= N ; ++i) {read(bx[i]);read(by[i]);read(bc[i]);}
    for(int i = 1 ; i <= N ; ++i) {
	addtwo(S,a[0][i],rc[i],0);
	for(int j = 0 ; j < 4 ; ++j) {
	    addtwo(a[0][i],ty[j],1e9,dx[j] * rx[i] + dy[j] * ry[i]);
	}
    }
    for(int i = 1 ; i <= N ; ++i) {
	addtwo(a[1][i],T,bc[i],0);
	for(int j = 0 ; j < 4 ; ++j) {
	    addtwo(ty[j],a[1][i],1e9,-dx[j] * bx[i] - dy[j] * by[i]);
	}
    }
}
void Solve() {
    int64 ans = 0;
    while(SPFA() > 0) {
	int p = preE[T],f = 1e9;
	while(p) {
	    f = min(f,E[p].cap);
	    p = preE[E[p ^ 1].to];
	}
	ans += dis[T] * f;
	p = preE[T];
	while(p) {
	    E[p].cap -= f;E[p ^ 1].cap += f;
	    p = preE[E[p ^ 1].to];
	}
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
}


### 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 <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 2005
#define ba 47
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 +c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
int S,T,Ncnt,N;
int a[2][MAXN],ty[4];
struct node {
    int to,next,cap;int64 val;
}E[MAXN * 100];
int head[MAXN * 2],sumE = 1;
int rx[MAXN],ry[MAXN],rc[MAXN];
int bx[MAXN],by[MAXN],bc[MAXN];
int dx[4] = {1,1,-1,-1};
int dy[4] = {1,-1,1,-1};

void add(int u,int v,int c,int64 a) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    E[sumE].cap = c;
    E[sumE].val = a;
    head[u] = sumE;
}
void addtwo(int u,int v,int c,int64 a) {
    add(u,v,c,a);
    add(v,u,0,-a);
}
int64 dis[MAXN];bool inq[MAXN];
int preE[MAXN];
queue<int> Q;
int64 SPFA() {
    for(int i = 1 ; i <= Ncnt ; ++i) dis[i] = -1e18;
    memset(inq,0,sizeof(inq));
    dis[S] = 0;inq[S] = 1;Q.push(S);
    while(!Q.empty()) {
	int u = Q.front();Q.pop();inq[u] = 0;
	for(int i = head[u] ; i ; i = E[i].next) {
	    int v = E[i].to;
	    if(E[i].cap) {
		if(dis[v] < dis[u] + E[i].val) {
		    dis[v] = dis[u] + E[i].val;
		    preE[v] = i;
		    if(!inq[v]) {Q.push(v);inq[v] = 1;}
		}
	    }
	}
    }
    return dis[T];
}
void Init() {
    read(N);
    S = ++Ncnt;
    for(int i = 0 ; i < 2 ; ++i) {
	for(int j = 1 ; j <= N ; ++j) {
	    a[i][j] = ++Ncnt;
	}
    }
    for(int i = 0 ; i < 4 ; ++i) ty[i] = ++Ncnt;
    T = ++Ncnt;
    for(int i = 1 ; i <= N ; ++i) {read(rx[i]);read(ry[i]);read(rc[i]);}
    for(int i = 1 ; i <= N ; ++i) {read(bx[i]);read(by[i]);read(bc[i]);}
    for(int i = 1 ; i <= N ; ++i) {
	addtwo(S,a[0][i],rc[i],0);
	for(int j = 0 ; j < 4 ; ++j) {
	    addtwo(a[0][i],ty[j],1e9,dx[j] * rx[i] + dy[j] * ry[i]);
	}
    }
    for(int i = 1 ; i <= N ; ++i) {
	addtwo(a[1][i],T,bc[i],0);
	for(int j = 0 ; j < 4 ; ++j) {
	    addtwo(ty[j],a[1][i],1e9,-dx[j] * bx[i] - dy[j] * by[i]);
	}
    }
}
void Solve() {
    int64 ans = 0;
    while(SPFA() > 0) {
	int p = preE[T],f = 1e9;
	while(p) {
	    f = min(f,E[p].cap);
	    p = preE[E[p ^ 1].to];
	}
	ans += dis[T] * f;
	p = preE[T];
	while(p) {
	    E[p].cap -= f;E[p ^ 1].cap += f;
	    p = preE[E[p ^ 1].to];
	}
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
}


```