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.*;
import java.util.*;


public class Main {
    static MyReader in = new MyReader();
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args){
        int N = in.i();
        Graph graph = new Graph(N, N - 1);
        LinkedList<Integer> list = new LinkedList<>();
        int[] color = new int[N + 1];
        for (int i = 1; i < color.length; i++) {
            color[i] = -1;
        }

        list.add(1);
        color[1] = 0;
        while (!list.isEmpty()) {
            int v = list.pop();
            int c = color[v];
            for (int[] a : graph.edge[v]) {
                int u = a[0];
                int w = a[1];
                if (color[u] == -1) {
                    color[u] = (w & 1) ^ c;
                    list.add(u);
                }
            }
        }

        for (int i = 1; i <= N; i++) {
            out.println(color[i]);
        }
        out.flush();
    }

    static class Graph {
        final int V;
        final int E;

        ArrayList<int[]>[] edge;

        Graph(int v, int e) {
            V = v;
            E = e;
            edge = new ArrayList[V + 1];
            for (int i = 1; i <= V; i++) {
                edge[i] = new ArrayList<>();
            }
            load();
        }

        void load() {
            for (int i = 0; i < E; i++) {
                int u = in.i();
                int v = in.i();
                int w = in.i();
                edge[u].add(new int[]{v, w});
                edge[v].add(new int[]{u, w});
            }
        }

    }

}

class MyReader extends BufferedReader {
    char[] cbuf = new char[1024];
    int head = 0;
    int tail = 0;

    MyReader() {
        super(new InputStreamReader(System.in));
    }

    char next() {
        if (head == tail) {
            try {
                tail = read(cbuf, 0, cbuf.length);
            } catch (IOException e) {}
            head = 0;
        }
        return cbuf[head++];
    }

    void back() {
        head--;
    }

    boolean minus() {
        boolean minus;
        while (true) {
            char c = next();
            if (c != ' ' && c != '\n' && c != '\r') {
                if (!(minus = c == '-')) back();
                return minus;
            }
        }
    }

    void skip() {
        while (true) {
            char c = next();
            if (c != ' ' && c != '\n' && c != '\r') {
                back();
                break;
            }
        }
    }

    char[] s(final int N) {
        skip();
        char[] s = new char[N];
        for (int i = 0; i < s.length; i++) {
            s[i] = next();
        }
        return s;
    }

    int i() {
        boolean minus = minus();
        int n = 0;
        while (true) {
            int k = next() - '0';
            if (k < 0 || 9 < k) break;
            n = 10 * n + k;
        }
        return minus ? -n : n;
    }

    int[] ii(final int N) {
        int[] a = new int[N];
        for (int j = 0; j < a.length; j++) a[j] = i();
        return a;
    }

    long l() {
        boolean minus = minus();
        long n = 0;
        while (true) {
            int k = next() - '0';
            if (k < 0 || 9 < k) break;
            n = 10 * n + k;
        }
        return minus ? -n : n;
    }
}