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) throws Exception {

        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        Node[] nodes = new Node[N];
        for(int i = 0; i < N; i++){
            nodes[i] = new Node(i);
        }
        
        for(int i = 0; i < N - 1; i++){
            int v1 = sc.nextInt() - 1;
            int v2 = sc.nextInt() - 1;
            int w  = sc.nextInt();
            
            nodes[v1].adjacents.put(nodes[v2], w);
            nodes[v2].adjacents.put(nodes[v1], w);
        }
        
        bfs(nodes[0]);
        
        for(Node n : nodes){
            System.out.println(n.color);
        }

    }
    
    public static void bfs(Node n0){
        
        Queue<Node> queue = new LinkedList<>();
        
        n0.color = 0;
        queue.add(n0);
        
        while(!queue.isEmpty()){
            
            Node n = queue.poll();
            
            for( Node na : n.adjacents.keySet() ){
                if( na.color != -1 ) continue;
                
                int na_kyori = n.adjacents.get(na);
                if ( na_kyori % 2 == 0 ){
                    na.color = n.color % 2;
                } else {
                    na.color = (n.color + 1) % 2;
                }
                
                queue.add(na);
                
            }
        }
    }
    
    static class Node{
        int id;
        // 白 : 0、 黒 : 1
        int color = -1;
        Map<Node, Integer> adjacents = new HashMap<>();
        
        public Node(int id){
            this.id = id;
        }
    }
}