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.Scanner;
import java.util.HashMap;
public class Main{
	static boolean[] kk;
	static HashMap<Integer,Long>[] map;
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		kk=new boolean[n];
		map=new HashMap[n];
		for(int i=0; i<n; i++){
			map[i]=new HashMap<>();
		}
		for(int i = 0; i<n-1; i++){
			int u = sc.nextInt();
			int v = sc.nextInt();
			long w = sc.nextLong();
			map[u-1].put(v,w);
			map[v-1].put(u,w);
		}
		dfs(-1,0,0);
		for(int i=0; i<n; i++){
			if(kk[i]){
				System.out.println(1);
			}else{
				System.out.println(0);
			}
		}
	}
	static void dfs(long sum,int r,int o){
		if(sum%2==0){
			kk[r]=true;
		}
		for(int j:map[r].keySet()){
			if(j-1!=o){
				dfs(map[r].get(j)+sum,j-1,r);
			}
		}
		return;
	}
}