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>

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define mkp make_pair
#define vi vector<int>
#define pii pair<int,int>
#define FI(n) FastIO::read(n)
#define FO(n) FastIO::write(n)
#define ull unsigned long long
#define mst(a,b) memset(a,b,sizeof(a))
#define foR(i,k,j) for(int i=(k);i>=(j);i--)
#define For(i,k,j) for(int i=(k);i<=(j);i++)
#define Foe(i,u) for(int i=lst[u],v=e[i].v;i;i=e[i].nxt,v=e[i].v)
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define Fin(s) freopen(s,"r",stdin)
#define Fout(s) freopen(s,"w",stdout)
#define file(s) Fin(s".in"),Fout(s".out")
#define INF ((1<<30)-1)
#define int long long
const int P=998244353; //
using namespace std;
template<typename T>inline void ckmax(T &a,T b) {(a<b)&&(a=b);}
template<typename T>inline void ckmin(T &a,T b) {(a>b)&&(a=b);}
inline int mul(int a,int b) {return 1ll*a*b%P;}
inline int add(int a,int b) {return a+b>=P?a+b-P:a+b;}
inline int sub(int a,int b) {return a-b>=0?a-b:a-b+P;}
inline void mulmod(int &a,int b) {a=mul(a, b);}
inline void addmod(int &a,int b) {((a+=b)>=P)&&(a-=P);}
inline void submod(int &a,int b) {((a-=b)<0)&&(a+=P);}
inline int ksm(int a,int b) {int ans=1; for(;b;b>>=1) {if(b&1) ans=1ll*ans*a%P;a=1ll*a*a%P;}return ans;}
inline void fprint(const vector<int> &f) {for(int i=0;i<(int)f.size();i++) fprintf(stderr,"%d ",f[i]); fprintf(stderr,"\n");}
inline int inv(int a) {return ksm(a,P-2);}
namespace FastIO {
  const int SIZE=1<<16; char buf[SIZE],obuf[SIZE],str[64]; int bi=SIZE,bn=SIZE,opt;
  int read(char *s) {
    while (bn) {for (;bi<bn&&buf[bi]<=' ';bi++);if (bi<bn) break; bn=fread(buf,1,SIZE,stdin),bi=0;}
    int sn=0;while (bn) {for (;bi<bn&&buf[bi]>' ';bi++) s[sn++]=buf[bi];if (bi<bn) break; bn=fread(buf,1,SIZE,stdin),bi=0;}s[sn]=0;return sn;
  }
  bool read(int& x) {if(x)x=0;int bf=0,n=read(str); if(!n) return 0; int i=0; if (str[i]=='-') bf=1,i=1; for(x=0;i<n;i++) x=x*10+str[i]-'0'; if(bf) x=-x; return 1;}
  void write(int x) {
    if(!x) obuf[opt++]='0'; else {if(x<0) obuf[opt++]='-',x=-x;int sn=0; while(x)str[sn++]=x%10+'0',x/=10;for (int i=sn-1;i>=0;i--) obuf[opt++]=str[i];}
    if (opt>=(SIZE>>1)){fwrite(obuf,1,opt,stdout); opt=0;}
  }
  void write(char x) {obuf[opt++]=x;if (opt>=(SIZE>>1)){fwrite(obuf,1,opt,stdout); opt=0;}}
	void Fflush() {if (opt) fwrite(obuf,1,opt,stdout); opt=0;}
};
inline int read() {int x; FI(x); return x;}
const int MN = 1e3 + 5;
struct Edge {
	int v, w, nxt;	
};
struct Dinic {
	static const int N = MN << 2;
	static const int M = MN << 4;
	int lst[N], tot, s, t, dis[N], cur[N];
	Edge e[M << 1];
	void init(int S, int T) {s = S, t = T, tot = 1; mst(lst, 0); }
	void addedge(int u, int v, int w) {
		e[++tot] = (Edge) {v, w, lst[u]}; lst[u] = tot;
		e[++tot] = (Edge) {u, 0, lst[v]}; lst[v] = tot;
	}
	bool bfs() {
		mst(dis, 0);
		queue < int > q; q.push(s); dis[s] = 1;
		while(q.size()) {
			int u = q.front(); q.pop();
			Foe(i, u) {
				if(e[i].w && !dis[v]) {
					dis[v] = dis[u] + 1;
					q.push(v);	
					if(v == t) return 1;
				}
			}
		}
		return dis[t];
	}
	int dfs(int u, int nowflow) {
		if(u == t) return nowflow;
		int flow = 0;
		for(int &i = cur[u]; i; i = e[i].nxt) {
			int v = e[i].v, di = 0;
			if(dis[v] == dis[u] + 1 && (di = dfs(v, min(nowflow, e[i].w))) ) 
				nowflow -= di, flow += di, e[i].w -= di, e[i ^ 1].w += di;
			if(!nowflow) break;
		}
		return flow;
	}
	int solve() {
		int ans = 0;
		while(bfs()) {
			memcpy(cur, lst, sizeof(cur));
			int di = 0;
			while((di = dfs(s, INF))) ans += di;
		}
		return ans;
	}
}d; 
int n, x[MN], y[MN], w[MN], ans;
#define in(i) (i * 2 - 1)
#define out(i) (i * 2)
map < pii, int > v;
int id(int i, int j) {
	
	if(i % 2 == 1 && j % 2 == 1) return 3;
	if(i % 2 == 1 && j % 2 == 0) return 2;
	if(i % 2 == 0 && j % 2 == 1) return 0;
	if(i % 2 == 0 && j % 2 == 0) return 1;
	assert(0); return -1;
}
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, -1, 1};
signed main() {
	#ifndef ONLINE_JUDGE
		freopen("pro.in","r",stdin);
		freopen("pro.out","w",stdout);
	#endif
	n = read();
	For(i, 1, n) {
		v[mkp(x[i] = read() + 1e9, y[i] = read() + 1e9)] = i;
		ans += (w[i] = read());
//		cerr << x[i] << ' ' << y[i] << ' ' << w[i] << endl; 
	}
	d.init(0, 2 * n + 1);
	For(i, 1, n) {
		d.addedge(in(i), out(i), w[i]);
	}
	For(i, 1, n) {
		For(k, 0, 3) {
			int tx = x[i] + dx[k], ty = y[i] + dy[k]; int j = v[mkp(tx, ty)];
			if(j && id(x[j], y[j]) == id(x[i], y[i]) + 1) {
				d.addedge(out(i), in(j), INF);
			}
		}
	}
	For(i, 1, n) {
		if(id(x[i], y[i]) == 0) d.addedge(0, in(i), INF);
		else if(id(x[i], y[i]) == 3) d.addedge(out(i), 2 * n + 1, INF);
	}
//	cerr << d.solve() << endl; 
	cout << ans - d.solve() << endl; 
	return FastIO::Fflush(),0;
}