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: Runtime Error)

import java.util.ArrayList;
import java.util.Scanner;

public class Main {

	static final int WHITE = 0;
	static final int BLACK = 1;
	static int[] color;

	private static boolean dfs2(int from, int to, ArrayList<ArrayList<Edge>> graph, int sum) {
		boolean flg = false;
		ArrayList<Edge> nextV = graph.get(from);
		for(int i = 0; i < nextV.size(); i++) {
			sum += nextV.get(i).weight;
			if(nextV.get(i).to == to) {
				if(sum % 2 == 0) {
					// OK
					flg = true;
				}else {
					// NG
					flg = false;
				}
			}else {
				// 探索を再帰的に呼び出す
				flg = dfs2(i, to, graph, sum);
			}
		}
		return flg;
	}


	private static boolean dfs(ArrayList<ArrayList<Edge>> graph, int v, int c){
		// vを探索済みにする
		color[v] = c;

		// 全ての頂点について調べる
		for(int i = 0; i < graph.size(); i++) {
			if(i == v || color[i] == -1) {
				continue;
			}
			// 同じ色に塗られた任意の2頂点について、その距離が偶数であるかを確認する
			if(color[i] == c) {
				if(dfs2(v, i, graph, 0)) {
					// OK
					continue;
				}else {
					// NG
					return false;
				}
			}
		}
		return true;

	}

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();

		ArrayList<ArrayList<Edge>> graph = new ArrayList<ArrayList<Edge>>();
		for(int i = 0; i < n + 1; i++) {
			graph.add(new ArrayList<Edge>());
		}

		int v1, v2, weight;
		// グラフ情報入力
		for(int i = 0; i < n - 1; i++) {
			v1 = scan.nextInt();
			v2 = scan.nextInt();
			weight = scan.nextInt();
			graph.get(v1).add(new Edge(v2,weight));
			graph.get(v2).add(new Edge(v1, weight));
		}

		scan.close();
		color = new int[n + 1];
		for(int i = 0; i < n; i++) {
			color[i] = -1;
		}

		// 探索
		for(int i = 1; i < n + 1; i++) {
			// iが探索済みであれば探索しない
			if(color[i] != -1) {
				continue;
			}
			if(dfs(graph, i, WHITE)) {
				// OK
			}else {
				dfs(graph, i, BLACK);
			}
		}
		for(int i = 1; i < n + 1; i++) {
			System.out.println(color[i]);
		}
	}
}
class Edge{
	int to;
	int weight;
	Edge(int to, int weight){
		this.to = to;
		this.weight = weight;
	}
}