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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;

public class Main {

    int n;
    E[] g;

    void run() {
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        g = new E[n];
        for (int i = 0; i < n; i++) g[i] = new E();

        for (int i = 0; i < n - 1; i++) {
            int u = sc.nextInt() - 1, v = sc.nextInt() - 1, w = sc.nextInt();
            g[u].add(new P(v, w));
            g[v].add(new P(u, w));
        }

        P st = dfs(0, new boolean[n]);
        P ed = dfs(st.t, new boolean[n]);

        boolean[] ans = new boolean[n];
        draw(ed.t, 0, new boolean[n], ans);

        PrintWriter out = new PrintWriter(System.out);
        for (int i = 0; i < n; i++) {
            out.println(ans[i] ? 0 : 1);
        }
        out.flush();
    }

    void draw(int p, long c, boolean[] visited, boolean[] color) {
        visited[p] = true;
        color[p] = c % 2 == 0;

        for (P next: g[p]) if (!visited[next.t]) {
            draw(next.t, c + next.w, visited, color);
        }
    }

    P dfs(int p, boolean[] visited) {
        visited[p] = true;

        P c = new P(p, 0);

        for (P next: g[p]) if (!visited[next.t]) {
            P tmp = dfs(next.t, visited);
            tmp.w += next.w;
            if (c.w < tmp.w) {
                c = tmp;
            }
        }

        return c;
    }

    class P implements Comparable<P> {
        int t;
        long w;
        P(int t, long w) {
            this.t = t;
            this.w = w;
        }

        @Override
        public int compareTo(P o) {
            return (w - o.w > 0) ? 1 : -1;
        }
    }

    class E extends ArrayList<P> {}

    void debug(Object...os) {
        System.err.println(Arrays.deepToString(os));
    }

    public static void main(String[] args) {
        new Main().run();
    }
}