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<cstdio>
int n;
struct node{
	int to,next,v;
}e[200001];
int ans[100001];
int head[100001],cnt;
inline void AddEdge(int u,int v,int val) {
	e[++cnt].to=v;
	e[cnt].v=val;
	e[cnt].next=head[u];
	head[u]=cnt;
}
inline void dfs(int now,int prt,int sum) {
	ans[now]=sum;
	for(int i=head[now];i;i=e[i].next) {
		int lv=e[i].to;
		if(lv!=prt) {
			if(e[i].v%2) dfs(lv,now,sum^1);
			else dfs(lv,now,sum);
		}
	}
}
int main() {
	scanf("%d",&n);
	for(int i=1,u,v,w;i<n;i++) {
		scanf("%d %d %d",&u,&v,&w);
		AddEdge(u,v,w);
		AddEdge(v,u,w);
	}
	dfs(1,0,0);
	for(int i=1;i<=n;i++) printf("%d\n",ans[i]);
	return 0;
}