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 {

	private static int[] a;
	private static boolean[] visited;
	private static List<Integer>[] g;
	private static Map<Integer, Map<Integer, Integer>> ws;


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

		g = new ArrayList[n];
		ws = new HashMap<>();

		for (int i = 0; i < g.length; i++) {
			g[i] = new ArrayList<>();
			ws.put(i, new HashMap<>());
		}

		//ws = new int[n][n];
		//for (int i = 0; i < ws.length; i++) {
		//	Arrays.fill(ws[i], Integer.MAX_VALUE);
		//}


		for (int i = 0; i < n - 1; i++) {
			int u = sc.nextInt() - 1;
			int v = sc.nextInt() - 1;
			int w = sc.nextInt() % 2;
			g[u].add(v);
			g[v].add(u);
			ws.get(u).put(v, w);
			ws.get(v).put(u, w);
			//ws[u][v] = w;
			//ws[v][u] = w;
		}

		a = new int[n];
		visited = new boolean[n];

		dfs(0, 0);
		for (int i : a) System.out.println(i);

		//int[] a = new int[n];
		//boolean[] visited = new boolean[n];
		//Queue<Integer> q = new LinkedList<>();
		//Queue<Integer> r = new LinkedList<>();
		//q.offer(0);
		//
		//while (!q.isEmpty()) {
		//	while (!q.isEmpty()) {
		//		int p = q.poll();
		//		if (visited[p]) continue;
		//		visited[p] = true;
		//		a[p] = 0;
		//
		//		for (Integer x : g[p]) {
		//			if (!visited[x]) {
		//				if (ws[p][x] == 0) {
		//					q.offer(x);
		//				} else {
		//					r.offer(x);
		//				}
		//			}
		//		}
		//	}
		//
		//	while (!r.isEmpty()) {
		//		int p = r.poll();
		//		if (visited[p]) continue;
		//		visited[p] = true;
		//		a[p] = 1;
		//
		//		for (Integer x : g[p]) {
		//			if (!visited[x]) {
		//				if (ws[p][x] == 0) {
		//					r.offer(x);
		//				} else {
		//					q.offer(x);
		//				}
		//			}
		//		}
		//	}
		//}

		//for (int i : a) {
		//	System.out.println(i);
		//}
	}

	private static void dfs(int v, int c) {
		if (visited[v]) return;
		visited[v] = true;
		a[v] = c;
		for (int i = 0; i < g[v].size(); i++) {
			int x = g[v].get(i);
			if (visited[x]) continue;
			if (ws.get(v).get(x) == 0) dfs(x, c);
			else dfs(x, c == 0 ? 1 : 0);
		}
	}
}