#include <bits/stdc++.h>

typedef long long LL;
const int N = 2015, M = 20005, inf = 0x3f3f3f3f;

int n, head[N], cur[N], tot = 1, s, t;
struct edge { int to, nxt, rem, cost; } e[M];

void link(int x, int y, int z, int w) {
	e[++tot] = (edge) { y, head[x], z, w }, head[x] = tot;
	e[++tot] = (edge) { x, head[y], 0, -w}, head[y] = tot;
}

LL ans = 0;
bool inq[N];
LL dis[N];

bool bellmanford() {
	std::memset(dis, -0x3f, sizeof dis);
	std::queue<int> q; q.push(s), inq[s] = 1, dis[s] = 0;
	while (q.size()) {
		int tmp = q.front(); q.pop(); inq[tmp] = 0;
		for (int i = head[tmp]; i; i = e[i].nxt)
			if (e[i].rem && dis[e[i].to] < dis[tmp] + e[i].cost) {
				dis[e[i].to] = dis[tmp] + e[i].cost;
				if (!inq[e[i].to]) inq[e[i].to] = 1, q.push(e[i].to);
			}
	}
	return dis[t] > -1e15;
}
bool vis[N];
int dinic(int x, int low) {
	if (x == t || !low) return low;
	int now = 0, t; vis[x] = 1;
	for (int &i = cur[x]; i; i = e[i].nxt)
		if (!vis[e[i].to] && e[i].rem && dis[e[i].to] == dis[x] + e[i].cost 
			&& (t = dinic(e[i].to, std::min(e[i].rem, low)))) {
				now += t, low -= t, e[i].rem -= t, e[i ^ 1].rem += t;
				ans += (LL) e[i].cost * t;
				if (!low) break;
			}
	vis[x] = 0;
	return now;
}


int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n, s = 0, t = n + n + 5;
	for (int i = 1; i <= n; ++i) {
		int x, y, c;
		std::cin >> x >> y >> c;
		link(s, i, c, 0);
		link(i, 2 * n + 1, c, x + y);
		link(i, 2 * n + 2, c, x - y);
		link(i, 2 * n + 3, c, -x + y);
		link(i, 2 * n + 4, c, -x - y);
	}
	for (int i = 1; i <= n; ++i) {
		int x, y, c;
		std::cin >> x >> y >> c;
		link(i + n, t, c, 0);
		link(2 * n + 1, i + n, c, -x - y);
		link(2 * n + 2, i + n, c, -x + y);
		link(2 * n + 3, i + n, c, x - y);
		link(2 * n + 4, i + n, c, x + y);
	}
	while (bellmanford()) std::memcpy(cur, head, sizeof head), dinic(s, inf);
	std::cout << ans << '\n';
	return 0;
}
### Prompt

Please formulate a CPP solution 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 <bits/stdc++.h>

typedef long long LL;
const int N = 2015, M = 20005, inf = 0x3f3f3f3f;

int n, head[N], cur[N], tot = 1, s, t;
struct edge { int to, nxt, rem, cost; } e[M];

void link(int x, int y, int z, int w) {
	e[++tot] = (edge) { y, head[x], z, w }, head[x] = tot;
	e[++tot] = (edge) { x, head[y], 0, -w}, head[y] = tot;
}

LL ans = 0;
bool inq[N];
LL dis[N];

bool bellmanford() {
	std::memset(dis, -0x3f, sizeof dis);
	std::queue<int> q; q.push(s), inq[s] = 1, dis[s] = 0;
	while (q.size()) {
		int tmp = q.front(); q.pop(); inq[tmp] = 0;
		for (int i = head[tmp]; i; i = e[i].nxt)
			if (e[i].rem && dis[e[i].to] < dis[tmp] + e[i].cost) {
				dis[e[i].to] = dis[tmp] + e[i].cost;
				if (!inq[e[i].to]) inq[e[i].to] = 1, q.push(e[i].to);
			}
	}
	return dis[t] > -1e15;
}
bool vis[N];
int dinic(int x, int low) {
	if (x == t || !low) return low;
	int now = 0, t; vis[x] = 1;
	for (int &i = cur[x]; i; i = e[i].nxt)
		if (!vis[e[i].to] && e[i].rem && dis[e[i].to] == dis[x] + e[i].cost 
			&& (t = dinic(e[i].to, std::min(e[i].rem, low)))) {
				now += t, low -= t, e[i].rem -= t, e[i ^ 1].rem += t;
				ans += (LL) e[i].cost * t;
				if (!low) break;
			}
	vis[x] = 0;
	return now;
}


int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n, s = 0, t = n + n + 5;
	for (int i = 1; i <= n; ++i) {
		int x, y, c;
		std::cin >> x >> y >> c;
		link(s, i, c, 0);
		link(i, 2 * n + 1, c, x + y);
		link(i, 2 * n + 2, c, x - y);
		link(i, 2 * n + 3, c, -x + y);
		link(i, 2 * n + 4, c, -x - y);
	}
	for (int i = 1; i <= n; ++i) {
		int x, y, c;
		std::cin >> x >> y >> c;
		link(i + n, t, c, 0);
		link(2 * n + 1, i + n, c, -x - y);
		link(2 * n + 2, i + n, c, -x + y);
		link(2 * n + 3, i + n, c, x - y);
		link(2 * n + 4, i + n, c, x + y);
	}
	while (bellmanford()) std::memcpy(cur, head, sizeof head), dinic(s, inf);
	std::cout << ans << '\n';
	return 0;
}
```