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) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		
		//方針：BFS
		//利用；キュー、隣接リスト
		
		List<ArrayList<int[]>> l = new ArrayList<ArrayList<int[]>>();
		for(int i=0;i<N;i++) l.add(new ArrayList<int[]>());
		
		for(int i=0;i<N-1;i++) {
			int from = sc.nextInt();
			int to = sc.nextInt();
			int dist = sc.nextInt();
			from--;
			to--;
			int tmp1[] = {to,dist};
			l.get(from).add(tmp1);
			int tmp2[] = {from,dist};
			l.get(to).add(tmp2);
		}
		
		int ans[] = new int[N];
		Arrays.fill(ans,-1);
		ans[0]=0;
		
		Queue<Integer> q = new ArrayDeque<Integer>();
		q.add(0);
		while(!q.isEmpty()) {
			int from = q.poll();
			List<int[]> tos = l.get(from);
			for(int tmp[] :tos) {
				int to = tmp[0];
				int dist = tmp[1];
				if(ans[to]!=-1)continue;
				ans[to]=(ans[from]+dist)%2;
				q.offer(to);
			}
		}
		
		StringBuilder sb = new StringBuilder();
		Arrays.stream(ans).forEach(i->sb.append(i + "\n"));
		System.out.println(sb.toString());
		
	}
}