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 pair<int,int> pi;
vector<pi>tree[100005];
bool visited[100005];
int color[100005];
void dfs(int curr){
	visited[curr] = true;
	for(int i=0; i<tree[curr].size(); i++){
		int nxt = tree[curr][i].first;
		int p = tree[curr][i].second;
		if(!visited[nxt]){
			if(p) color[nxt] = !color[curr];
			else color[nxt] = color[curr];
			dfs(nxt);
		}
	}
}
int main(){
	int n;
	cin>>n;
	for(int i=1; i<n; i++){
		int a,b,w;
		cin>>a>>b>>w;
		tree[a].push_back(pi(b,w%2));
		tree[b].push_back(pi(a,w%2));
	}
	dfs(1);
	for(int i=1; i<=n; i++) cout<<color[i]<<'\n';
}