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;
typedef long long ll;
int main() {
	ll n;
	cin>>n;
	vector<ll>ans(n+1,-1);
	ans[1]=0;
	vector<vector<pair<ll,ll>>>road(n+1);
	for(int i=0;i<n-1;i++){
		ll u,v,w;
		cin>>u>>v>>w;
		road[u].push_back({v,w%2});
		road[v].push_back({u,w%2});
	}
	queue<ll>que;
	que.push(1);
	while(!que.empty()){
		ll x=que.front();
		que.pop();
		for(pair<ll,ll>p:road[x]){
			if(ans[p.first]!=-1)continue;
			ans[p.first]=(ans[x]+p.second)%2;
			que.push(p.first);
		}
	}
	for(int i=1;i<=n;i++)cout<<ans[i]<<endl;
	return 0;
}