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<bits/stdc++.h>
using namespace std;
int n,t1,t2,t3;
bool color[100005],vis[100005];
vector<pair<int,bool> > a[100005];
void dfs(int x,bool c){
	if(vis[x]){
		return;
	}
	color[x]=c;
	vis[x]=1;
	for(int i=0;i<a[x].size();i++){
		if(a[x][i].second){
			dfs(a[x][i].first,!c);
		}else{
			dfs(a[x][i].first,c);
		}
	}
}
int main(){
	cin>>n;
	for(int i=1;i<n;i++){
		cin>>t1>>t2>>t3;
		if(t3%2==0){
			a[t1].push_back({t2,0});
			a[t2].push_back({t1,0});
		}else{
			a[t1].push_back({t2,1});
			a[t2].push_back({t1,1});
		}
	}
	dfs(1,0);
	for(int i=1;i<=n;i++){
		cout<<color[i]<<endl;
	}
	return 0;
}