#include <cstdlib>
#include <cstdio>
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <ctime>
#include <algorithm>
#define N 5678
#define NN 401000
typedef long long ll;
template <typename T> inline void read(T &x) {
	x = 0; char c = getchar(); bool flag = false;
	while (!isdigit(c)) { if (c == '-')	flag = true; c = getchar(); }
	while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); }
	if (flag)	x = -x;
}
using namespace std;
const int inf = 987654321;
int n, s, t;
struct edge{
	int nxt, to, val;
	ll fare;
}e[NN];
int head[N], ecnt = 1, hcur[N];
inline void addedge(int from, int to, int val, ll fare) {
	e[++ecnt] = (edge){head[from], to, val, fare};
	head[from] = ecnt;
}
inline void Addedge(int u, int v, int w, ll f) {
	addedge(u, v, w, -f);
	addedge(v, u, 0, f);
}

int que[NN * 30], front, rear;
ll dis[N];
bool vis[N];
inline bool spfa() {
	front = rear = 0;
	memset(dis, 0x3f, sizeof(dis));
	for (int i = 1; i <= t; ++i)	hcur[i] = head[i];
	
	que[++rear] = s; dis[s] = 0;
	while (front < rear) {
		int cur = que[++front]; vis[cur] = false;
		for (int i = head[cur]; i; i = e[i].nxt) {
			int to = e[i].to;
			if (!e[i].val || dis[to] <= dis[cur] + e[i].fare)	continue;
			dis[to] = dis[cur] + e[i].fare;
			if (!vis[to])	vis[to] = true, que[++rear] = to;
		}
	}
	return dis[t] <= 1e14;
}
int dfs(int cur, int limi) {
	if (cur == t || !limi)	return limi;
	int flow = 0;
	vis[cur] = true;
	for (int i = hcur[cur]; i; i = e[i].nxt) {
		int to = e[i].to; hcur[cur] = i;
		if (vis[to] || dis[to] != dis[cur] + e[i].fare)	 continue;
		int tmp = dfs(to, min(e[i].val, limi));
		if (!tmp)	continue;
		flow += tmp; limi -= tmp;
		e[i].val -= tmp; e[i ^ 1].val += tmp;
		if (!limi)	break;
	}
	vis[cur] = false;
	return flow;
}

int dx[4] = {1, -1, 1, -1};
int dy[4] = {-1, 1, 1, -1};
int main() {
	read(n);
	s = n + n + 5, t = s + 1;
	for (int i = 1; i <= n; ++i) {
		int x, y, ct; read(x), read(y), read(ct);
		for (int tp = 0; tp < 4; ++tp) {
			int xx = x * dx[tp], yy = y * dy[tp];
			Addedge(i, n + n + tp + 1, ct, xx + yy);
		}
		Addedge(s, i, ct, 0);
	}
	for (int i = 1; i <= n; ++i) {
		int x, y, ct; read(x), read(y), read(ct);
		for (int tp = 0; tp < 4; ++tp) {
			int xx = x * dx[tp] * (-1), yy = y * dy[tp] * (-1);
			Addedge(n + n + tp + 1, n + i, ct, xx + yy);
		}
		Addedge(n + i, t, ct, 0);
	}
	int mxflow = 0;
	ll res = 0;
	while (spfa()) {
		int tmp = dfs(s, inf);
		mxflow += tmp;
		res += dis[t] * tmp;
	}
	printf("%lld\n", -res);
	return 0;
}

/*
2
0 0 1
3 2 1
2 2 1
5 0 1
//8
*/
### Prompt

Please create a solution in cpp to 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 <cstdlib>
#include <cstdio>
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <ctime>
#include <algorithm>
#define N 5678
#define NN 401000
typedef long long ll;
template <typename T> inline void read(T &x) {
	x = 0; char c = getchar(); bool flag = false;
	while (!isdigit(c)) { if (c == '-')	flag = true; c = getchar(); }
	while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); }
	if (flag)	x = -x;
}
using namespace std;
const int inf = 987654321;
int n, s, t;
struct edge{
	int nxt, to, val;
	ll fare;
}e[NN];
int head[N], ecnt = 1, hcur[N];
inline void addedge(int from, int to, int val, ll fare) {
	e[++ecnt] = (edge){head[from], to, val, fare};
	head[from] = ecnt;
}
inline void Addedge(int u, int v, int w, ll f) {
	addedge(u, v, w, -f);
	addedge(v, u, 0, f);
}

int que[NN * 30], front, rear;
ll dis[N];
bool vis[N];
inline bool spfa() {
	front = rear = 0;
	memset(dis, 0x3f, sizeof(dis));
	for (int i = 1; i <= t; ++i)	hcur[i] = head[i];
	
	que[++rear] = s; dis[s] = 0;
	while (front < rear) {
		int cur = que[++front]; vis[cur] = false;
		for (int i = head[cur]; i; i = e[i].nxt) {
			int to = e[i].to;
			if (!e[i].val || dis[to] <= dis[cur] + e[i].fare)	continue;
			dis[to] = dis[cur] + e[i].fare;
			if (!vis[to])	vis[to] = true, que[++rear] = to;
		}
	}
	return dis[t] <= 1e14;
}
int dfs(int cur, int limi) {
	if (cur == t || !limi)	return limi;
	int flow = 0;
	vis[cur] = true;
	for (int i = hcur[cur]; i; i = e[i].nxt) {
		int to = e[i].to; hcur[cur] = i;
		if (vis[to] || dis[to] != dis[cur] + e[i].fare)	 continue;
		int tmp = dfs(to, min(e[i].val, limi));
		if (!tmp)	continue;
		flow += tmp; limi -= tmp;
		e[i].val -= tmp; e[i ^ 1].val += tmp;
		if (!limi)	break;
	}
	vis[cur] = false;
	return flow;
}

int dx[4] = {1, -1, 1, -1};
int dy[4] = {-1, 1, 1, -1};
int main() {
	read(n);
	s = n + n + 5, t = s + 1;
	for (int i = 1; i <= n; ++i) {
		int x, y, ct; read(x), read(y), read(ct);
		for (int tp = 0; tp < 4; ++tp) {
			int xx = x * dx[tp], yy = y * dy[tp];
			Addedge(i, n + n + tp + 1, ct, xx + yy);
		}
		Addedge(s, i, ct, 0);
	}
	for (int i = 1; i <= n; ++i) {
		int x, y, ct; read(x), read(y), read(ct);
		for (int tp = 0; tp < 4; ++tp) {
			int xx = x * dx[tp] * (-1), yy = y * dy[tp] * (-1);
			Addedge(n + n + tp + 1, n + i, ct, xx + yy);
		}
		Addedge(n + i, t, ct, 0);
	}
	int mxflow = 0;
	ll res = 0;
	while (spfa()) {
		int tmp = dfs(s, inf);
		mxflow += tmp;
		res += dis[t] * tmp;
	}
	printf("%lld\n", -res);
	return 0;
}

/*
2
0 0 1
3 2 1
2 2 1
5 0 1
//8
*/
```