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 CPP):

Solution:
#include<stdio.h>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
int n,xx,yy,zz,b[110000],be[110000],et;
struct edg{
	int y,z,ne;
}e[210000];
inline void add_edge(int x,int y,int z){
	e[++et].y=y;
	e[et].z=z;
	e[et].ne=be[x];
	be[x]=et;
}
void dfs(int x,int fa){
	for(int i=be[x];i;i=e[i].ne) if (e[i].y!=fa){
		if (e[i].z&1) b[e[i].y]=b[x]^1;else b[e[i].y]=b[x];
		dfs(e[i].y,x);
	}
}
int main(){
	scanf("%d",&n);
	fo(i,2,n){
		scanf("%d%d%d",&xx,&yy,&zz);
		add_edge(xx,yy,zz);
		add_edge(yy,xx,zz);
	}
	dfs(1,0);
	fo(i,1,n) printf("%d\n",b[i]);
	return 0;
}