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 int MAX=1e+5;
vector<int> g[MAX];
map<int,map<int,int>> d;
int check[MAX]={0};bool color[MAX];
void dfs(int v){
	int u;
	for(int i=0;i<g[v].size();i++){
		u=g[v][i];
		if(check[u]==0){
			check[u]++;
			if(d[v][u]==0){
				color[u]=color[v];
			}else{
				color[u]=!color[v];
			}dfs(u);
		}
	}return;
}
int main(){
	int N;cin>>N;
    int u,v,w;
	for(int i=0;i<N-1;i++){
		cin>>u>>v>>w;
		u--;v--;w=w%2;
		g[u].push_back(v);
		g[v].push_back(u);
		d[u][v]=w;
		d[v][u]=w;
	}check[0]++;color[0]=true;
	dfs(0);
	for(int j=0;j<N;j++){
		cout<<color[j]<<endl;
	}
}