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()
{
	int n;
	cin >> n;
	vector<vector<pair<int,int>>> M(n);
	for(int i =0;i<n-1;i++)
	{
		int u,v,w;
		cin >> u >> v >> w;
		u--;
		v--;
		M[u].emplace_back(v,w);
		M[v].emplace_back(u,w);
	}
	vector<int> color(n,-1);
	stack<int> s;
	s.push(0);
	color[0] = 0;
	while (!s.empty())
	{
		int now = s.top();
		s.pop();
		for(auto i:M[now])
		{
			if(color[i.first]==-1)
			{
				s.push(i.first);
				color[i.first] = (color[now]+i.second+2)%2;
			}
		}
	}
	for(auto i:color)cout<<i<<endl;
	return 0;
}