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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.Queue;

public class Main {

	static int ans[];

	public static void main(String args[]) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		int n = Integer.parseInt(br.readLine());
		ans = new int[n];

		ArrayList<ArrayList<Edge>> list = new ArrayList<>();

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

		for (int i = 0; i < n - 1; i++) {
			String str[] = br.readLine().split(" ");
			int u = Integer.parseInt(str[0]) - 1;
			int v = Integer.parseInt(str[1]) - 1;
			int w = Integer.parseInt(str[2]);

			list.get(u).add(new Edge(v, w));
			list.get(v).add(new Edge(u, w));
		}

		dfs(0, -1, 0, list);
		for (int i = 0; i < n; i++) {
			System.out.println(ans[i]);
		}
	}

	static void dfs(int v, int parent, int color, ArrayList<ArrayList<Edge>> edge) {
		ans[v] = color;

		for (Edge e : edge.get(v)) {
			if (e.to == parent)
				continue;
			if (e.cost % 2 == 0) {
				dfs(e.to, v, color, edge);
			} else {
				dfs(e.to, v, 1 - color, edge);
			}
		}
	}
}

class Edge implements Comparable<Edge> {
	int to;
	int cost;

	public Edge(int to, int cost) {
		this.to = to;
		this.cost = cost;
	}

	@Override
	public int compareTo(Edge e) {
		return this.cost - e.cost;
	}
}