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

public class Main {

	static class Edge {
		int to, cost;

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

	static List<Edge>[] g;
	static int[] color;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int N = sc.nextInt(); // 頂点数
		g = new ArrayList[N + 1]; // 辺の情報
		color = new int[N + 1]; // 塗る色

		for (int i = 0; i < N + 1; i++) {
			g[i] = new ArrayList<Edge>();
		}

		for (int i = 0; i < N - 1; i++) { // 辺はN-1本
			int u = sc.nextInt();
			int v = sc.nextInt();
			int w = sc.nextInt();
			g[u].add(new Edge(v, w));
			g[v].add(new Edge(u, w));
		}

		dfs(1, 0, 0);

		for (int i = 1; i < N + 1; i++) {
			System.out.println(color[i]);
		}

		sc.close();
	}

	public static void dfs(int u, int parent, long sum) {
		color[u] = (int) (sum % 2l);
		for (Edge e : g[u]) {
			if (e.to == parent) {
				continue;
			}
			dfs(e.to, u, sum + e.cost);
		}
	}

}