Problem: 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.

Examples

Input

3
1 2 2
2 3 1


Output

0
0
1


Input

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


Output

1
0
1
0
1

Solution (in JAVA):

Solution:

import java.util.*;

public class Main {
	public static void main(String[] args) {
		Main main = new Main();
		main.run();
	}

	class Pair{
		int to;
		int w;
		Pair(int to, int w){
			this.to=to;
			this.w=w;
		}
	}

	int N;
	ArrayList<ArrayList<Pair>> g;
	public void run() {
		Scanner sc = new Scanner(System.in);
		N=sc.nextInt();
		g = new ArrayList<>();
		for(int i=0; i<N; i++) {
			g.add(new ArrayList<Pair>());
		}
		for(int i=0; i<N-1; i++) {
			int v1 = sc.nextInt();
			int v2 = sc.nextInt();
			v1--;
			v2--;
			int w = sc.nextInt();
			g.get(v1).add(new Pair(v2, w));
			g.get(v2).add(new Pair(v1, w));
		}
		bfs(0);
		sc.close();
	}

	private void bfs(int r) {
		boolean used[] = new boolean[N];
		int color[] = new int[N];
		ArrayDeque<Integer> q = new ArrayDeque<Integer>();
		q.add(r);
		used[r]=true;
		color[r]=1;
		long ans[] = new long[N];
		ans[r]=0;

		while(!q.isEmpty()) {
			int u = q.poll();
			for(int i=0; i<g.get(u).size(); i++) {
				Pair p = g.get(u).get(i);
				if(used[p.to]) continue;
				ans[p.to]=ans[u]+p.w;
				used[p.to]=true;
				q.add(p.to);
			}
		}

		for(int i=0; i<N; i++) {
			System.out.println(ans[i]%2);
		}
	}
}