#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

ll const inf = 1e18;

struct MCMF{
	int sr, sn;
	vector<ll> d;
	vector<int> dr;
	vector<vector<int> > g;
	
	struct ED{
		int v, cp;
		ll cs, cc;
		ED(){}
		ED(int a, int b, ll c):v(a), cp(b), cs(c), cc(c) {}
	};
	
	vector<ED> ed;
	
	pair<int, ll> kl(int v, int cp = 1e9){
		if (v == sn)return make_pair(cp, 0);
		int fl = 0;
		ll cs = 0;
		for (int t: g[v])if (ed[t].cp){
			int u = ed[t].v;
			if (dr[v] + 1 != dr[u] || ed[t].cc)continue;
			pair<int, ll> an = kl(u, min(ed[t].cp, cp - fl));
			fl += an.first;
			cs += ed[t].cs * an.first + an.second;
			ed[t].cp -= an.first;
			ed[t ^ 1].cp += an.first;
			if (fl == cp)break;
		}
		if (fl < cp)dr[v] = 0;
		return make_pair(fl, cs);
	}
	
	void upd() { f(i, 0, g.size())for (int t: g[i])ed[t].cc += d[i] - d[ed[t].v]; }
	
	void di(){
		f(i, 0, d.size())d[i] = inf;
		d[sr] = 0;
		f(i, 0, dr.size())dr[i] = 0;
		priority_queue<pair<ll, int> > q;
		q.push(make_pair(0, sr));
		while (!q.empty()){
			ll c = -q.top().first;
			int v = q.top().second;
			q.pop();
			if (c > d[v])continue;
			for (int t: g[v])if (ed[t].cp){
				int u = ed[t].v;
				ll nc = c + ed[t].cc;
				if (nc < d[u])dr[u] = dr[v] + 1, q.push(make_pair(-(d[u] = nc), u));
			}
		}
		upd();
	}
	
	pair<int, ll> mcmf(int s, int t){
		sr = s, sn = t;
		d = vector<ll>(g.size(), inf);
		dr = vector<int>(g.size(), 1e9);
		d[sr] = dr[sr] = 0;
		f(it, 1, g.size())f(i, 0, g.size())if (d[i] != inf)for (int t: g[i])if (ed[t].cp){
			int u = ed[t].v;
			ll nc = d[i] + ed[t].cs;
			if (nc < d[u])d[u] = nc, dr[u] = dr[i] + 1;
		}
		upd();
		int fl = 0;
		ll cs = 0;
		while (d[sn] != inf){
			pair<int, ll> t = kl(sr);
			fl += t.first;
			cs += t.second;
			di();
		}
		return make_pair(fl, cs);
	}
	
	void aded(int a, int b, int w1 = 1, int cst = 0, int w2 = 0){
		g.resize(max((int)g.size(), max(a, b) + 1));
		g[a].push_back(ed.size());
		ed.emplace_back(b, w1, cst);
		g[b].push_back(ed.size());
		ed.emplace_back(a, w2, -cst);
	}
};

int const NN = 2000;
pair<int, int> p[NN];
int dp[NN], n, tp[4];
 
int main(){
	scanf("%d", &n);
	f(i, 0, n)scanf("%d%d%d", &p[i].first, &p[i].second, dp + i);
	f(i, 0, n)scanf("%d%d%d", &p[i + n].first, &p[i + n].second, dp + i + n);
	n <<= 1;
	int sr = n, sn = n | 1;
	f(i, 0, 4)tp[i] = n + 2 + i;
	n >>= 1;
	MCMF mf;
	f(i, 0, n)mf.aded(sr, i, dp[i]);
	f(i, n, n << 1)mf.aded(i, sn, dp[i]);
	f(i, 0, n)f(j, 0, 4)mf.aded(i, tp[j], 1e9, ((j & 1) ? p[i].first : -p[i].first) + ((j & 2) ? p[i].second : -p[i].second));
	f(i, n, n << 1)f(j, 0, 4)mf.aded(tp[j], i, 1e9, ((j & 1) ? -p[i].first : p[i].first) + ((j & 2) ? -p[i].second : p[i].second));
	printf("%lld\n", -mf.mcmf(sr, sn).second);
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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>
using namespace std;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

ll const inf = 1e18;

struct MCMF{
	int sr, sn;
	vector<ll> d;
	vector<int> dr;
	vector<vector<int> > g;
	
	struct ED{
		int v, cp;
		ll cs, cc;
		ED(){}
		ED(int a, int b, ll c):v(a), cp(b), cs(c), cc(c) {}
	};
	
	vector<ED> ed;
	
	pair<int, ll> kl(int v, int cp = 1e9){
		if (v == sn)return make_pair(cp, 0);
		int fl = 0;
		ll cs = 0;
		for (int t: g[v])if (ed[t].cp){
			int u = ed[t].v;
			if (dr[v] + 1 != dr[u] || ed[t].cc)continue;
			pair<int, ll> an = kl(u, min(ed[t].cp, cp - fl));
			fl += an.first;
			cs += ed[t].cs * an.first + an.second;
			ed[t].cp -= an.first;
			ed[t ^ 1].cp += an.first;
			if (fl == cp)break;
		}
		if (fl < cp)dr[v] = 0;
		return make_pair(fl, cs);
	}
	
	void upd() { f(i, 0, g.size())for (int t: g[i])ed[t].cc += d[i] - d[ed[t].v]; }
	
	void di(){
		f(i, 0, d.size())d[i] = inf;
		d[sr] = 0;
		f(i, 0, dr.size())dr[i] = 0;
		priority_queue<pair<ll, int> > q;
		q.push(make_pair(0, sr));
		while (!q.empty()){
			ll c = -q.top().first;
			int v = q.top().second;
			q.pop();
			if (c > d[v])continue;
			for (int t: g[v])if (ed[t].cp){
				int u = ed[t].v;
				ll nc = c + ed[t].cc;
				if (nc < d[u])dr[u] = dr[v] + 1, q.push(make_pair(-(d[u] = nc), u));
			}
		}
		upd();
	}
	
	pair<int, ll> mcmf(int s, int t){
		sr = s, sn = t;
		d = vector<ll>(g.size(), inf);
		dr = vector<int>(g.size(), 1e9);
		d[sr] = dr[sr] = 0;
		f(it, 1, g.size())f(i, 0, g.size())if (d[i] != inf)for (int t: g[i])if (ed[t].cp){
			int u = ed[t].v;
			ll nc = d[i] + ed[t].cs;
			if (nc < d[u])d[u] = nc, dr[u] = dr[i] + 1;
		}
		upd();
		int fl = 0;
		ll cs = 0;
		while (d[sn] != inf){
			pair<int, ll> t = kl(sr);
			fl += t.first;
			cs += t.second;
			di();
		}
		return make_pair(fl, cs);
	}
	
	void aded(int a, int b, int w1 = 1, int cst = 0, int w2 = 0){
		g.resize(max((int)g.size(), max(a, b) + 1));
		g[a].push_back(ed.size());
		ed.emplace_back(b, w1, cst);
		g[b].push_back(ed.size());
		ed.emplace_back(a, w2, -cst);
	}
};

int const NN = 2000;
pair<int, int> p[NN];
int dp[NN], n, tp[4];
 
int main(){
	scanf("%d", &n);
	f(i, 0, n)scanf("%d%d%d", &p[i].first, &p[i].second, dp + i);
	f(i, 0, n)scanf("%d%d%d", &p[i + n].first, &p[i + n].second, dp + i + n);
	n <<= 1;
	int sr = n, sn = n | 1;
	f(i, 0, 4)tp[i] = n + 2 + i;
	n >>= 1;
	MCMF mf;
	f(i, 0, n)mf.aded(sr, i, dp[i]);
	f(i, n, n << 1)mf.aded(i, sn, dp[i]);
	f(i, 0, n)f(j, 0, 4)mf.aded(i, tp[j], 1e9, ((j & 1) ? p[i].first : -p[i].first) + ((j & 2) ? p[i].second : -p[i].second));
	f(i, n, n << 1)f(j, 0, 4)mf.aded(tp[j], i, 1e9, ((j & 1) ? -p[i].first : p[i].first) + ((j & 2) ? -p[i].second : p[i].second));
	printf("%lld\n", -mf.mcmf(sr, sn).second);
}
```