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 <iostream>
#include <vector>
using namespace std;
const int MAXN=100005;
int n;
vector<pair<int,int> > v[MAXN];
int color[MAXN];

void dfs(int no,int father)
{
	for (int i=0;i<v[no].size();i++){
		int u=v[no][i].first;
		int w=v[no][i].second%2;
		if (u!=father){
			color[u]=color[no]^w;
			dfs(u,no);
		}
	}
}

int main()
{
	cin>>n;
	for (int i=1;i<n;i++){
		int u,vv,c;
		cin>>u>>vv>>c;
		v[u].push_back({vv,c});
		v[vv].push_back({u,c});
	}

	dfs(1,-1);

	for (int i=1;i<=n;i++){
		cout<<color[i]<<endl;
	}
	return 0;
}