Problem:
Score : 400 points

Problem Statement

We have a tree with N vertices numbered 1 to N.
The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i.
Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied:

For any two vertices painted in the same color, the distance between them is an even number.

Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq u_i < v_i \leq N

1 \leq w_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
u_1 v_1 w_1
u_2 v_2 w_2
.
.
.
u_{N - 1} v_{N - 1} w_{N - 1}

Output

Print a coloring of the vertices that satisfies the condition, in N lines.
The i-th line should contain 0 if Vertex i is painted white and 1 if it is painted black.

If there are multiple colorings that satisfy the condition, any of them will be accepted.

Sample Input 1

3
1 2 2
2 3 1

Sample Output 1

0
0
1

Sample Input 2

5
2 5 2
2 3 10
1 3 8
3 4 2

Sample Output 2

1
0
1
0
1

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

//#pragma once
#include <stdio.h>
#include <string>
#include <iostream>
#include <queue>
#include <algorithm>
#include <sstream>
#include <vector>
#include <math.h>
#include <set>
#include <numeric>

using namespace std;

typedef pair<long long, long long> pl;

#define ll long long


long long A[200000] = {};

class UnionFind {
public:
	vector <ll> par; 
	vector <ll> siz;
	vector <ll> gr;

	// Constructor
	UnionFind(ll sz_) : par(sz_), siz(sz_, 1LL) ,gr(sz_){
		for (ll i = 0; i < sz_; ++i) par[i] = i; 
	}
	void init(ll sz_) {
		par.resize(sz_);
		siz.assign(sz_, 1LL);  // resize だとなぜか初期化されなかった
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
	}

	// Member Function
	// Find
	ll root(ll x) { // 根の検索
		while (par[x] != x) {
			x = par[x] = par[par[x]]; // x の親の親を x の親とする
		}
		return x;
	}

	// Union(Unite, Merge)
	bool merge(ll x, ll y) {
		x = root(x);
		y = root(y);
		if (x == y) return false;

		// merge technique（データ構造をマージするテク．小を大にくっつける）
		if (siz[x] < siz[y]) swap(x, y);
		siz[x] += siz[y];
		par[y] = x;
		return true;
	}



	bool issame(ll x, ll y) { // 連結判定
		return root(x) == root(y);
	}

	ll size(ll x) { // 素集合のサイズ
		return siz[root(x)];
	}

	bool mergeA(ll x, ll y, ll z, ll n) {
		/*x = root(x);
		y = root(y);
		if (x == y) return false;*/
		if (z % 2 == 1 && A[x] % 2 == A[y] % 2) {
			for (int i = 0; i < n; i++) {
				if (par[i]==par[x]) {
					A[i]++;
				}
			}
		}

		if (z % 2 == 0 && A[x] % 2 != A[y] % 2) {
			for (int i = 0; i < n; i++) {
				if (par[i] == par[x]) {
					A[i]++;
				}
			}
		}
		// merge technique（データ構造をマージするテク．小を大にくっつける）
		if (siz[x] < siz[y]) swap(x, y);
		siz[x] += siz[y];
		par[y] = x;
		return true;
	}

};


long long N, M, k, a, b, c, d, e, H, W;
long long B[3][100005] = {};
bool f;
bool ff;
string S;
set <long long>sll;
pl buf;
vector <long long>vl;

struct ST
{
	long long group;
	long long rank;
	long long tugi=-1;
	int siro;
};

vector <ST> vst;
ST st[100004];

double fl, fll;


int main() {

	cin >> N ;

	UnionFind uf(N);

	for (int i = 0; i < N-1; i++) {
		cin >> B[0][i] >> B[1][i] >> B[2][i];
		
		uf.mergeA(B[0][i]-1, B[1][i]-1,B[2][i],N);
	

	}

	for (int i = 0; i < N; i++) {
		cout << A[i] % 2 << endl;
	}

	return 0;

}