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;
const long long max_n=100010;
long long n,c[max_n];
vector<pair<long long,long long> > t[max_n];
void dfs(long long b,long long previous=-1){
	for (auto it: t[b]){
		long long a=it.first;
		long long d=it.second%2;
		if (a!=previous){
			c[a]=c[b]^d;
			dfs(a,b);
		}
	}
}
int main()
{
	cin>>n;
	for (int i=0;i<n-1;++i){
		long long a,b,d;
		cin>>a>>b>>d;
		t[a].push_back({b,d});
		t[b].push_back({a,d}); 
	}
	dfs(1);
	for (int i=1;i<=n;++i)
	    cout<<c[i]<<endl;
}