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;

public class Main {

    public static void main(String[] args) {
        Scanner std = new Scanner(System.in);
        int N = std.nextInt();

        List<List<Integer>> to = new ArrayList<>(N);
        List<List<Integer>> cost = new ArrayList<>(N);
        for (int i = 0; i < N; i++) {
            to.add(new ArrayList<>());
            cost.add(new ArrayList<>());
        }

        for (int i = 0; i < N - 1; i++) {
            int a = std.nextInt();
            int b = std.nextInt();
            int w = std.nextInt();
            a--;
            b--;

            to.get(a).add(b);
            to.get(b).add(a);
            cost.get(a).add(w);
            cost.get(b).add(w);
        }

        int[] ans = new int[N];
        Deque<Integer> deque = new ArrayDeque<>();
        Arrays.fill(ans, - 1);
        ans[0] = 0;
        deque.add(0);
        while (!deque.isEmpty()) {
            int v = deque.pollFirst();
            for (int i = 0; i < to.get(v).size(); i++) {
                int u = to.get(v).get(i);
                int w = cost.get(v).get(i);
                if (ans[u] != -1) {
                    continue;
                }

                ans[u] = (ans[v] + w) % 2;
                deque.add(u);
            }
        }

        Arrays.stream(ans).map(Math::abs).forEach(System.out::println);
    }
}