#include <iostream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <string>
#include <algorithm>
#include <utility>
#include <complex>
#define rep(x, s, t) for(llint (x) = (s); (x) <= (t); (x)++)
#define per(x, s, t) for(llint (x) = (s); (x) >= (t); (x)--)
#define reps(x, s) for(llint (x) = 0; (x) < (llint)(s).size(); (x)++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))
#define sz(x) ((ll)(x).size())
#define ceil(x, y) (((x)+(y)-1) / (y))
#define all(x) (x).begin(),(x).end()
#define outl(...) dump_func(__VA_ARGS__)
#define inf 1e18

using namespace std;

typedef long long llint;
typedef long long ll;
typedef pair<ll, ll> P;

struct edge{
	ll to, cost;
	edge(){}
	edge(ll a, ll b){ to = a, cost = b;}
};
const ll dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};

//const ll mod = 1000000007;
const ll mod = 998244353;

struct mint{
	ll x = 0;
	mint(ll y = 0){x = y; if(x < 0 || x >= mod) x = (x%mod+mod)%mod;}
	mint(const mint &ope) {x = ope.x;}
	
	mint operator-(){return mint(-x);}
	mint operator+(const mint &ope){return mint(x) += ope;}
	mint operator-(const mint &ope){return mint(x) -= ope;}
	mint operator*(const mint &ope){return mint(x) *= ope;}
	mint operator/(const mint &ope){return mint(x) /= ope;}
	mint& operator+=(const mint &ope){
		x += ope.x;
		if(x >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint &ope){
		x += mod - ope.x;
		if(x >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint &ope){
		x *= ope.x, x %= mod;
		return *this;
	}
	mint& operator/=(const mint &ope){
		ll n = mod-2; mint mul = ope;
		while(n){
			if(n & 1) *this *= mul;
			mul *= mul;
			n >>= 1;
		}
		return *this;
	}
	mint inverse(){return mint(1) / *this;}
	bool operator ==(const mint &ope){return x == ope.x;}
	bool operator !=(const mint &ope){return x != ope.x;}
};
mint modpow(mint a, ll n){
	if(n == 0) return mint(1);
	if(n % 2) return a * modpow(a, n-1);
	else return modpow(a*a, n/2);
}
istream& operator >>(istream &is, mint &ope){
	ll t; is >> t, ope.x = t;
	return is;
}
ostream& operator <<(ostream &os, mint &ope){return os << ope.x;}
ostream& operator <<(ostream &os, const mint &ope){return os << ope.x;}

bool exceed(ll x, ll y, ll m){return x >= m / y + 1;}
void mark(){ cout << "*" << endl; }
void yes(){ cout << "YES" << endl; }
void no(){ cout << "NO" << endl; }
ll gcd(ll a, ll b){if(b == 0) return a; return gcd(b, a%b);}
ll digitnum(ll x, ll b = 10){ll ret = 0; for(; x; x /= b) ret++; return ret;}
ll digitsum(ll x, ll b = 10){ll ret = 0; for(; x; x /= b) ret += x % b; return ret;}
string lltos(ll x){string ret; for(;x;x/=10) ret += x % 10 + '0'; reverse(ret.begin(), ret.end()); return ret;}
ll stoll(string &s){ll ret = 0; for(auto c : s) ret *= 10, ret += c - '0'; return ret;}
template<typename T>
void uniq(T &vec){ sort(vec.begin(), vec.end()); vec.erase(unique(vec.begin(), vec.end()), vec.end());}

template<typename T>
ostream& operator << (ostream& os, vector<T>& vec) {
	for(int i = 0; i < vec.size(); i++) os << vec[i] << (i + 1 == vec.size() ? "" : " ");
	return os;
}
template<typename T>
ostream& operator << (ostream& os, deque<T>& deq) {
	for(int i = 0; i < deq.size(); i++) os << deq[i] << (i + 1 == deq.size() ? "" : " ");
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, pair<T, U>& pair_var) {
	os << "(" << pair_var.first << ", " << pair_var.second << ")";
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, const pair<T, U>& pair_var) {
	os << "(" << pair_var.first << ", " << pair_var.second << ")";
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, map<T, U>& map_var) {
	for(typename map<T, U>::iterator itr = map_var.begin(); itr != map_var.end(); itr++) {
		os << "(" << itr->first << ", " << itr->second << ")";
		itr++; if(itr != map_var.end()) os << ","; itr--;
	}
	return os;
}
template<typename T>
ostream& operator << (ostream& os, set<T>& set_var) {
	for(typename set<T>::iterator itr = set_var.begin(); itr != set_var.end(); itr++) {
		os << *itr; ++itr; if(itr != set_var.end()) os << " "; itr--;
	}
	return os;
}
template<typename T>
ostream& operator << (ostream& os, multiset<T>& set_var) {
	for(typename multiset<T>::iterator itr = set_var.begin(); itr != set_var.end(); itr++) {
		os << *itr; ++itr; if(itr != set_var.end()) os << " "; itr--;
	}
	return os;
}
template<typename T>
void outa(T a[], ll s, ll t){
	for(ll i = s; i <= t; i++){ cout << a[i]; if(i < t) cout << " ";}
	cout << endl;
}
void dump_func() {cout << endl;}
template <class Head, class... Tail>
void dump_func(Head &&head, Tail &&... tail) {
	cout << head;
	if(sizeof...(Tail) > 0) cout << " ";
	dump_func(std::move(tail)...);
}


struct Dinic{
	struct edge{
		llint to, cap, rev;
		edge(llint a, llint b, llint c){
			to = a, cap = b, rev = c;
		}
	};
	
	int n;
	vector<vector<edge> > G;
	vector<llint> level, iter;
	Dinic(){}
	Dinic(int n){
		this->n = n;
		G.resize(n+1);
		level.resize(n+1);
		iter.resize(n+1);
	}
	void init(){
		for(int i = 0; i <= n; i++) G[i].clear();
	}
	void add_edge(int s, int t, llint cap)
	{
		G[s].push_back(edge(t, cap, G[t].size()));
		G[t].push_back(edge(s, 0, G[s].size()-1));
	}
	void bfs(int s)
	{
		for(int i = 0; i <= n; i++) level[i] = inf;
		level[s] = 0;
		
		queue<int> Q;
		Q.push(s);
		while(Q.size()){
			int v = Q.front();
			Q.pop();
			for(int i = 0; i < G[v].size(); i++){
				int u = G[v][i].to;
				if(G[v][i].cap <= 0 || level[u] < inf) continue;
				level[u] = level[v] + 1;
				Q.push(u);
			}
		}
	}
	llint dfs(int v, llint f, int t)
	{
		if(v == t) return f;
		
		llint ret;
		for(llint &i = iter[v]; i < G[v].size(); i++){
			if(level[v] >= level[G[v][i].to] || G[v][i].cap <= 0) continue;
			ret = dfs(G[v][i].to, min(f, G[v][i].cap), t);
			if(ret > 0){
				G[v][i].cap -= ret;
				G[G[v][i].to][G[v][i].rev].cap += ret;
				return ret;
			}
		}
		return 0;
	}
	llint calc(int s, int t)
	{
		llint ret = 0, flow;
		while(1){
			bfs(s);
			if(level[t] >= inf) break;
			for(int i = 0; i <= n; i++) iter[i] = 0;
			while(1){
				flow = dfs(s, inf, t);
				if(flow <= 0) break;
				ret += flow;
			}
		}
		return ret;
	}
};

ll n;
ll x[1005], y[1005], w[1005];
map<P, ll> mp;
Dinic dc(2005);

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n;
	rep(i, 1, n){
		cin >> x[i] >> y[i] >> w[i];
		mp[P(x[i], y[i])] = i;
	}
	ll ans = 0;
	rep(i, 1, n) ans += w[i];
	
	ll S = 2*n+1, T = 2*n+2;
	rep(i, 1, n) dc.add_edge(i, n+i, w[i]);
	rep(i, 1, n){
		if(x[i] % 2 && y[i] % 2 == 0){
			dc.add_edge(S, i, inf);
			if(mp.count(P(x[i]+1, y[i]))) dc.add_edge(n+i, mp[P(x[i]+1, y[i])], inf);
			if(mp.count(P(x[i]-1, y[i]))) dc.add_edge(n+i, mp[P(x[i]-1, y[i])], inf);
		}
		if(x[i] % 2 == 0 && y[i] % 2 == 0){
			if(mp.count(P(x[i], y[i]+1))) dc.add_edge(n+i, mp[P(x[i], y[i]+1)], inf);
			if(mp.count(P(x[i], y[i]-1))) dc.add_edge(n+i, mp[P(x[i], y[i]-1)], inf);
		}
		if(x[i] % 2 == 0 && y[i] % 2){
			if(mp.count(P(x[i]+1, y[i]))) dc.add_edge(n+i, mp[P(x[i]+1, y[i])], inf);
			if(mp.count(P(x[i]-1, y[i]))) dc.add_edge(n+i, mp[P(x[i]-1, y[i])], inf);
		}
		if(x[i] % 2 && y[i] % 2) dc.add_edge(n+i, T, inf);
	}
	ans -= dc.calc(S, T);
	
	outl(ans);
	
	return 0;
}
### Prompt

Generate a Cpp solution to the following problem:

At the foot of Liyushan Mountain, n tents will be carefully arranged to provide accommodation for those who are willing to experience the joy of approaching nature, the tranquility of the night, and the bright starry sky.

The i-th tent is located at the point of (x_i, y_i) and has a weight of w_i. A tent is important if and only if both x_i and y_i are even. You need to remove some tents such that for each remaining important tent (x, y), there do not exist 3 other tents (x'_1, y'_1), (x'_2, y'_2) and (x'_3, y'_3) such that both conditions are true: 

  1. |x'_j-x|, |y'_j - y|≤ 1 for all j ∈ \{1, 2, 3\}, and 
  2. these four tents form a parallelogram (or a rectangle) and one of its sides is parallel to the x-axis. 



Please maximize the sum of the weights of the tents that are not removed. Print the maximum value.

Input

The first line contains a single integer n (1≤ n≤ 1 000), representing the number of tents.

Each of the next n lines contains three integers x_i, y_i and w_i (-10^9≤ x_i,y_i ≤ 10^9, 1≤ w_i≤ 10^9), representing the coordinate of the i-th tent and its weight. No two tents are located at the same point.

Output

A single integer — the maximum sum of the weights of the remaining tents.

Examples

Input


5
0 0 4
0 1 5
1 0 3
1 1 1
-1 1 2


Output


12


Input


32
2 2 1
2 3 1
3 2 1
3 3 1
2 6 1
2 5 1
3 6 1
3 5 1
2 8 1
2 9 1
1 8 1
1 9 1
2 12 1
2 11 1
1 12 1
1 11 1
6 2 1
7 2 1
6 3 1
5 3 1
6 6 1
7 6 1
5 5 1
6 5 1
6 8 1
5 8 1
6 9 1
7 9 1
6 12 1
5 12 1
6 11 1
7 11 1


Output


24

Note

Here is an illustration of the second example. Black triangles indicate the important tents. This example also indicates all 8 forbidden patterns.

<image>

### Response

```cpp
#include <iostream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <string>
#include <algorithm>
#include <utility>
#include <complex>
#define rep(x, s, t) for(llint (x) = (s); (x) <= (t); (x)++)
#define per(x, s, t) for(llint (x) = (s); (x) >= (t); (x)--)
#define reps(x, s) for(llint (x) = 0; (x) < (llint)(s).size(); (x)++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))
#define sz(x) ((ll)(x).size())
#define ceil(x, y) (((x)+(y)-1) / (y))
#define all(x) (x).begin(),(x).end()
#define outl(...) dump_func(__VA_ARGS__)
#define inf 1e18

using namespace std;

typedef long long llint;
typedef long long ll;
typedef pair<ll, ll> P;

struct edge{
	ll to, cost;
	edge(){}
	edge(ll a, ll b){ to = a, cost = b;}
};
const ll dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};

//const ll mod = 1000000007;
const ll mod = 998244353;

struct mint{
	ll x = 0;
	mint(ll y = 0){x = y; if(x < 0 || x >= mod) x = (x%mod+mod)%mod;}
	mint(const mint &ope) {x = ope.x;}
	
	mint operator-(){return mint(-x);}
	mint operator+(const mint &ope){return mint(x) += ope;}
	mint operator-(const mint &ope){return mint(x) -= ope;}
	mint operator*(const mint &ope){return mint(x) *= ope;}
	mint operator/(const mint &ope){return mint(x) /= ope;}
	mint& operator+=(const mint &ope){
		x += ope.x;
		if(x >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint &ope){
		x += mod - ope.x;
		if(x >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint &ope){
		x *= ope.x, x %= mod;
		return *this;
	}
	mint& operator/=(const mint &ope){
		ll n = mod-2; mint mul = ope;
		while(n){
			if(n & 1) *this *= mul;
			mul *= mul;
			n >>= 1;
		}
		return *this;
	}
	mint inverse(){return mint(1) / *this;}
	bool operator ==(const mint &ope){return x == ope.x;}
	bool operator !=(const mint &ope){return x != ope.x;}
};
mint modpow(mint a, ll n){
	if(n == 0) return mint(1);
	if(n % 2) return a * modpow(a, n-1);
	else return modpow(a*a, n/2);
}
istream& operator >>(istream &is, mint &ope){
	ll t; is >> t, ope.x = t;
	return is;
}
ostream& operator <<(ostream &os, mint &ope){return os << ope.x;}
ostream& operator <<(ostream &os, const mint &ope){return os << ope.x;}

bool exceed(ll x, ll y, ll m){return x >= m / y + 1;}
void mark(){ cout << "*" << endl; }
void yes(){ cout << "YES" << endl; }
void no(){ cout << "NO" << endl; }
ll gcd(ll a, ll b){if(b == 0) return a; return gcd(b, a%b);}
ll digitnum(ll x, ll b = 10){ll ret = 0; for(; x; x /= b) ret++; return ret;}
ll digitsum(ll x, ll b = 10){ll ret = 0; for(; x; x /= b) ret += x % b; return ret;}
string lltos(ll x){string ret; for(;x;x/=10) ret += x % 10 + '0'; reverse(ret.begin(), ret.end()); return ret;}
ll stoll(string &s){ll ret = 0; for(auto c : s) ret *= 10, ret += c - '0'; return ret;}
template<typename T>
void uniq(T &vec){ sort(vec.begin(), vec.end()); vec.erase(unique(vec.begin(), vec.end()), vec.end());}

template<typename T>
ostream& operator << (ostream& os, vector<T>& vec) {
	for(int i = 0; i < vec.size(); i++) os << vec[i] << (i + 1 == vec.size() ? "" : " ");
	return os;
}
template<typename T>
ostream& operator << (ostream& os, deque<T>& deq) {
	for(int i = 0; i < deq.size(); i++) os << deq[i] << (i + 1 == deq.size() ? "" : " ");
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, pair<T, U>& pair_var) {
	os << "(" << pair_var.first << ", " << pair_var.second << ")";
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, const pair<T, U>& pair_var) {
	os << "(" << pair_var.first << ", " << pair_var.second << ")";
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, map<T, U>& map_var) {
	for(typename map<T, U>::iterator itr = map_var.begin(); itr != map_var.end(); itr++) {
		os << "(" << itr->first << ", " << itr->second << ")";
		itr++; if(itr != map_var.end()) os << ","; itr--;
	}
	return os;
}
template<typename T>
ostream& operator << (ostream& os, set<T>& set_var) {
	for(typename set<T>::iterator itr = set_var.begin(); itr != set_var.end(); itr++) {
		os << *itr; ++itr; if(itr != set_var.end()) os << " "; itr--;
	}
	return os;
}
template<typename T>
ostream& operator << (ostream& os, multiset<T>& set_var) {
	for(typename multiset<T>::iterator itr = set_var.begin(); itr != set_var.end(); itr++) {
		os << *itr; ++itr; if(itr != set_var.end()) os << " "; itr--;
	}
	return os;
}
template<typename T>
void outa(T a[], ll s, ll t){
	for(ll i = s; i <= t; i++){ cout << a[i]; if(i < t) cout << " ";}
	cout << endl;
}
void dump_func() {cout << endl;}
template <class Head, class... Tail>
void dump_func(Head &&head, Tail &&... tail) {
	cout << head;
	if(sizeof...(Tail) > 0) cout << " ";
	dump_func(std::move(tail)...);
}


struct Dinic{
	struct edge{
		llint to, cap, rev;
		edge(llint a, llint b, llint c){
			to = a, cap = b, rev = c;
		}
	};
	
	int n;
	vector<vector<edge> > G;
	vector<llint> level, iter;
	Dinic(){}
	Dinic(int n){
		this->n = n;
		G.resize(n+1);
		level.resize(n+1);
		iter.resize(n+1);
	}
	void init(){
		for(int i = 0; i <= n; i++) G[i].clear();
	}
	void add_edge(int s, int t, llint cap)
	{
		G[s].push_back(edge(t, cap, G[t].size()));
		G[t].push_back(edge(s, 0, G[s].size()-1));
	}
	void bfs(int s)
	{
		for(int i = 0; i <= n; i++) level[i] = inf;
		level[s] = 0;
		
		queue<int> Q;
		Q.push(s);
		while(Q.size()){
			int v = Q.front();
			Q.pop();
			for(int i = 0; i < G[v].size(); i++){
				int u = G[v][i].to;
				if(G[v][i].cap <= 0 || level[u] < inf) continue;
				level[u] = level[v] + 1;
				Q.push(u);
			}
		}
	}
	llint dfs(int v, llint f, int t)
	{
		if(v == t) return f;
		
		llint ret;
		for(llint &i = iter[v]; i < G[v].size(); i++){
			if(level[v] >= level[G[v][i].to] || G[v][i].cap <= 0) continue;
			ret = dfs(G[v][i].to, min(f, G[v][i].cap), t);
			if(ret > 0){
				G[v][i].cap -= ret;
				G[G[v][i].to][G[v][i].rev].cap += ret;
				return ret;
			}
		}
		return 0;
	}
	llint calc(int s, int t)
	{
		llint ret = 0, flow;
		while(1){
			bfs(s);
			if(level[t] >= inf) break;
			for(int i = 0; i <= n; i++) iter[i] = 0;
			while(1){
				flow = dfs(s, inf, t);
				if(flow <= 0) break;
				ret += flow;
			}
		}
		return ret;
	}
};

ll n;
ll x[1005], y[1005], w[1005];
map<P, ll> mp;
Dinic dc(2005);

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n;
	rep(i, 1, n){
		cin >> x[i] >> y[i] >> w[i];
		mp[P(x[i], y[i])] = i;
	}
	ll ans = 0;
	rep(i, 1, n) ans += w[i];
	
	ll S = 2*n+1, T = 2*n+2;
	rep(i, 1, n) dc.add_edge(i, n+i, w[i]);
	rep(i, 1, n){
		if(x[i] % 2 && y[i] % 2 == 0){
			dc.add_edge(S, i, inf);
			if(mp.count(P(x[i]+1, y[i]))) dc.add_edge(n+i, mp[P(x[i]+1, y[i])], inf);
			if(mp.count(P(x[i]-1, y[i]))) dc.add_edge(n+i, mp[P(x[i]-1, y[i])], inf);
		}
		if(x[i] % 2 == 0 && y[i] % 2 == 0){
			if(mp.count(P(x[i], y[i]+1))) dc.add_edge(n+i, mp[P(x[i], y[i]+1)], inf);
			if(mp.count(P(x[i], y[i]-1))) dc.add_edge(n+i, mp[P(x[i], y[i]-1)], inf);
		}
		if(x[i] % 2 == 0 && y[i] % 2){
			if(mp.count(P(x[i]+1, y[i]))) dc.add_edge(n+i, mp[P(x[i]+1, y[i])], inf);
			if(mp.count(P(x[i]-1, y[i]))) dc.add_edge(n+i, mp[P(x[i]-1, y[i])], inf);
		}
		if(x[i] % 2 && y[i] % 2) dc.add_edge(n+i, T, inf);
	}
	ans -= dc.calc(S, T);
	
	outl(ans);
	
	return 0;
}
```