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.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Deque;
import java.util.List;
import java.util.Scanner;

//方針はBFS
public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        List<List<Integer>> edge = new ArrayList<>(N+1);
        List<List<Integer>> cost = new ArrayList<>(N+1);
        for(int i=1;i<=N+1;i++){
          edge.add(new ArrayList<>());
          cost.add(new ArrayList<>());
        }

        for(int i=1;i<=N;i++){
          edge.get(i).add(0);
          cost.get(i).add(0);
        }

        for(int i=1;i<=N-1;i++){
          int u = sc.nextInt();
          int v = sc.nextInt();
          int w = sc.nextInt();

            edge.get(u).add(v);
            edge.get(v).add(u);
            cost.get(u).add(w);
            cost.get(v).add(w);

        }


        Deque <Integer> dq = new ArrayDeque<>();
        int status [] = new int[N+1];
        int ans [] = new int[N+1];


        for(int i=1;i<=N;i++){
            ans[i]=0;
            status[i]=0;
        }

        dq.push(1);
        status[1]=1;
        while(!dq.isEmpty()){
            int first_in = dq.pollLast();
            //
            for(int i=1;i<=edge.get(first_in).size()-1;i++){
                if(status[edge.get(first_in).get(i)]==0){
                    dq.push(edge.get(first_in).get(i));
                    status[edge.get(first_in).get(i)]=1;
                    int w = cost.get(first_in).get(i);
                    ans[edge.get(first_in).get(i)]=(ans[first_in] + w) % 2;
                    }
                }
      }
        for(int j=1;j<=N;j++) {
            System.out.println(ans[j]);
        }
    }
}