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.ArrayDeque;
import java.util.ArrayList;

public class Main {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] inputs = br.readLine().split(" ");
		int n = Integer.parseInt(inputs[0]);

		Node[] nodes = new Node[n];
		for ( int i = 0; i < n; i++ ) {
			nodes[i] = new Node();
		}

		for ( int i = 0; i < n - 1; i++ ) {
			inputs = br.readLine().split(" ");
			int u = Integer.parseInt(inputs[0]) - 1;
			int v = Integer.parseInt(inputs[1]) - 1;
			int w = Integer.parseInt(inputs[2]);
			nodes[u].nexts.add(new Edge(v, w));
			nodes[v].nexts.add(new Edge(u, w));
		}
		br.close();

		ArrayDeque<Integer> q = new ArrayDeque<>();
		q.add(0);
		int[] tree = new int[n];
		tree[0] = 1;
		while ( !q.isEmpty() ) {
			int idx = q.poll();

			for ( Edge e : nodes[idx].nexts ) {
				if ( tree[e.to] == 0 ) {
					q.add(e.to);

					if ( e.dist % 2 == 0 ) tree[e.to] = tree[idx];
					else tree[e.to] = tree[idx] * (-1);
				}
			}
		}

		for ( int i : tree ) {
			System.out.println(Math.max(i, 0));
		}
	}
}

class Node {

	ArrayList<Edge> nexts = new ArrayList<>();
}

class Edge {

	int	to		= 0;
	int	dist	= 0;

	public Edge(int t, int d) {
		to = t;
		dist = d;
	}
}