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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;


import java.io.*;

public class Main {
	static boolean[] c;
	static Map<Integer,Long>[] ma;
	@SuppressWarnings("unchecked")
	private static void solve(){
		//Implement solution here.
		int n = ni();
		c = new boolean[n + 1];
		ma = new HashMap[n + 1]; 
		for(int i = 0; i < n + 1; i++) {
			ma[i] = new HashMap<Integer,Long>();
		}

		for(int i = 0; i < n - 1; i++) {
			int u = ni();
			int v = ni();
			long w = nl();
			ma[u].put(v, w);
			ma[v].put(u, w);
		}
		
		dfs(0,1,0);
		for(int i = 1; i < n + 1; i++) {
			System.out.println(c[i] ? "0":"1");
		}

	}
	
	static void dfs(long wsum, int target, int before) {
		if(wsum %2 == 0) {
			c[target] = true;
		}
		for(int next:ma[target].keySet()) {
			if(next != before) {
				dfs(wsum + ma[target].get(next), next, target);
			}
		}		
	}
	

	//Switch input source (stdin/file)
	private static BufferedReader br  = new BufferedReader(new InputStreamReader(System.in));
	public static void main(String[] args){
		String debugDataPath = System.getenv("DD_PATH");        
		if(debugDataPath != null){
			try{
				br = new BufferedReader(new InputStreamReader(new FileInputStream(new File(debugDataPath))));
			}catch(Exception e){
				throw new RuntimeException(e);
			}
		} 
		solve();
	}

	//Input read utility
	private static StringTokenizer tokenizer = null;
	public static String next() {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}
	//Get next single int
	private static int ni() {
		return Integer.parseInt(next());
	}
	//Get next single long
	private static long nl() {
		return Long.parseLong(next());
	}
	//Get next single double
	private static double nd() {
		return Double.parseDouble(next());
	}
	//Get next int array from one line
	private static int[] na(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}

	//Get next int array from one line
	private static Integer[] nao(int n) {
		Integer[] a = new Integer[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}
	//Get next char array from one line
	private static char[] ns() {
		return next().toCharArray();
	}
	//Get next long array from one line
	private static long[] nal(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nl();
		return a;
	}
}