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 {
  static int n;
  static ArrayList[] e;
  static HashMap[] elen;
  static int[] c;
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    n = sc.nextInt();
    e = new ArrayList[n];
    elen = new HashMap[n];
    c = new int[n];
    for(int i = 0; i < n; i++) {
      e[i] = new ArrayList<Integer>();
      elen[i] = new HashMap<Integer, Integer>();
    }
    for(int i = 0; i < n - 1; i++) {
      int u = sc.nextInt() - 1;
      int v = sc.nextInt() - 1;
      int w = sc.nextInt();
      e[u].add(v);
      e[v].add(u);
      elen[u].put(v, w);
      elen[v].put(u, w);
    }
    dfs(0, -1);
    for(int i = 0; i < n; i++) {
      System.out.println(c[i]);
    }
  }

  public static void dfs(int now, int p) {
    ArrayList<Integer> l = e[now];
    HashMap<Integer, Integer> h = elen[now];
    for(int i = 0; i < l.size(); i++) {
      int g = l.get(i);
      if(g != p) {
        int len = h.get(g);
        if(len % 2 == 0) {
          c[g] = c[now];
        } else {
          c[g] = 1 - c[now];
        }
        dfs(g, now);
      }
    }
  }
}