#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#include <algorithm>
#define int long long
using namespace std;
const int N = 1e6 + 5, inf = 1e18;
int read() {
	int x = 0, f = 1; char ch;
	while(! isdigit(ch = getchar())) (ch == '-') && (f = -f);
	for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 3) + (x << 1) + (ch ^ 48));
	return x * f;
}
template <class T> T Max(T a, T b) { return a > b ? a : b; }
template <class T> T Min(T a, T b) { return a < b ? a : b; }
int n, s, t, p1, p2, p3, p4, mc, mf, cnt = 1, head[N], to[N], val[N], vis[N], dis[N], nxt[N], cost[N], flow[N], last[N], pre[N];
void add(int x, int y, int w, int c) {
	to[++ cnt] = y; val[cnt] = w; cost[cnt] = c; nxt[cnt] = head[x]; head[x] = cnt;
	to[++ cnt] = x; val[cnt] = 0; cost[cnt] = - c; nxt[cnt] = head[y]; head[y] = cnt;
}
queue <int> q;
bool spfa() {
	for(int i = s; i <= t; ++ i) dis[i] = -inf, flow[i] = inf;
	q.push(s); vis[s] = 1; dis[s] = 0; pre[t] = -1;
	while(! q.empty()) {
		int x = q.front(); q.pop(); vis[x] = 0;
		for(int i = head[x]; i; i = nxt[i]) {
			if(val[i] && dis[to[i]] < dis[x] + cost[i]) {
				dis[to[i]] = dis[x] + cost[i];
				flow[to[i]] = min(flow[x], val[i]);
				pre[to[i]] = x; last[to[i]] = i;
				if(! vis[to[i]]) vis[to[i]] = 1, q.push(to[i]);
			}
		}
	}
	return pre[t] != -1;
}
void mcmf() {
	while(spfa()) {
		int now = t; mc += flow[t] * dis[t];
		while(now != s) {
			val[last[now]] -= flow[t];
			val[last[now] ^ 1] += flow[t];
			now = pre[now];
		}
	}
}
signed main() {
	n = read(); s = 0; p1 = 2 * n + 1; p2 = p1 + 1; p3 = p2 + 1; p4 = p3 + 1; t = p4 + 1;
	for(int i = 1, x, y, z; i <= n; ++ i) {
		x = read(); y = read(); z = read();
		add(s, i, z, 0);
		add(i, p1, inf, x + y); add(i, p2, inf, x - y);
		add(i, p3, inf, - x + y); add(i, p4, inf, - x - y); 
	}
	for(int i = 1, x, y, z; i <= n; ++ i) {
		x = read(); y = read(); z = read();
		add(i + n, t, z, 0);
		add(p1, i + n, inf, - x - y); add(p2, i + n, inf, - x + y);
		add(p3, i + n, inf, x - y); add(p4, i + n, inf, x + y);
	}
	mcmf();
	printf("%lld\n", mc);
	return 0;
}

### Prompt

Your task is to create 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 <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#include <algorithm>
#define int long long
using namespace std;
const int N = 1e6 + 5, inf = 1e18;
int read() {
	int x = 0, f = 1; char ch;
	while(! isdigit(ch = getchar())) (ch == '-') && (f = -f);
	for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 3) + (x << 1) + (ch ^ 48));
	return x * f;
}
template <class T> T Max(T a, T b) { return a > b ? a : b; }
template <class T> T Min(T a, T b) { return a < b ? a : b; }
int n, s, t, p1, p2, p3, p4, mc, mf, cnt = 1, head[N], to[N], val[N], vis[N], dis[N], nxt[N], cost[N], flow[N], last[N], pre[N];
void add(int x, int y, int w, int c) {
	to[++ cnt] = y; val[cnt] = w; cost[cnt] = c; nxt[cnt] = head[x]; head[x] = cnt;
	to[++ cnt] = x; val[cnt] = 0; cost[cnt] = - c; nxt[cnt] = head[y]; head[y] = cnt;
}
queue <int> q;
bool spfa() {
	for(int i = s; i <= t; ++ i) dis[i] = -inf, flow[i] = inf;
	q.push(s); vis[s] = 1; dis[s] = 0; pre[t] = -1;
	while(! q.empty()) {
		int x = q.front(); q.pop(); vis[x] = 0;
		for(int i = head[x]; i; i = nxt[i]) {
			if(val[i] && dis[to[i]] < dis[x] + cost[i]) {
				dis[to[i]] = dis[x] + cost[i];
				flow[to[i]] = min(flow[x], val[i]);
				pre[to[i]] = x; last[to[i]] = i;
				if(! vis[to[i]]) vis[to[i]] = 1, q.push(to[i]);
			}
		}
	}
	return pre[t] != -1;
}
void mcmf() {
	while(spfa()) {
		int now = t; mc += flow[t] * dis[t];
		while(now != s) {
			val[last[now]] -= flow[t];
			val[last[now] ^ 1] += flow[t];
			now = pre[now];
		}
	}
}
signed main() {
	n = read(); s = 0; p1 = 2 * n + 1; p2 = p1 + 1; p3 = p2 + 1; p4 = p3 + 1; t = p4 + 1;
	for(int i = 1, x, y, z; i <= n; ++ i) {
		x = read(); y = read(); z = read();
		add(s, i, z, 0);
		add(i, p1, inf, x + y); add(i, p2, inf, x - y);
		add(i, p3, inf, - x + y); add(i, p4, inf, - x - y); 
	}
	for(int i = 1, x, y, z; i <= n; ++ i) {
		x = read(); y = read(); z = read();
		add(i + n, t, z, 0);
		add(p1, i + n, inf, - x - y); add(p2, i + n, inf, - x + y);
		add(p3, i + n, inf, x - y); add(p4, i + n, inf, x + y);
	}
	mcmf();
	printf("%lld\n", mc);
	return 0;
}

```