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();
    List<List<Integer>> to = new ArrayList<>();
    List<List<Integer>> cost = new ArrayList<>();
    for (int i = 0; i < N; i++) {
      to.add(new ArrayList<>());
      cost.add(new ArrayList<>());
    }
    for (int i = 0; i < N-1; i++) {
      int u = sc.nextInt();
      int v = sc.nextInt();
      int w = sc.nextInt();
      u--;
      v--;
      to.get(u).add(v);
      to.get(v).add(u);
      cost.get(u).add(w);
      cost.get(v).add(w);
    }
    
    // BFS
    int[] ans = new int[N];
    Arrays.fill(ans, -1);
    LinkedList<Integer> q = new LinkedList<>();
    q.offer(0);
    ans[0] = 0;
    while (!q.isEmpty()) {
      int v = q.poll();
      for (int i = 0; i < to.get(v).size(); i++) {
        int vv = to.get(v).get(i);
        if (ans[vv] == -1) {
          ans[vv] = cost.get(v).get(i)%2 == 0 ? ans[v] : ans[v]^1;
          q.offer(vv);
        }
      }
    }
    
    for (int i = 0; i < N; i++) {
      System.out.println(ans[i]);
    }
  }
}