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;
#define rep(i,n) for(int i=0; i<(n); i++)

typedef long long ll;
typedef pair<ll,int> P;

int main(){
	int N;
	cin >> N;
	vector<vector<int>> G(N);
	map<P,int> d;
	rep(i,N-1){
		int u,v,w;
		cin >> u >> v >> w;
		u--; v--; w%=2;
		d[P(u,v)] = d[P(v,u)] = w;
		G[u].push_back(v);
		G[v].push_back(u);
	}

	vector<int> used(N,-1);
	used[0] = 0;
	queue<int> que;
	que.push(0);
	while(!que.empty()){
		int x = que.front(); que.pop();
		for(int u : G[x]){
			if(used[u]==-1){
				used[u] = used[x] xor d[P(x,u)];
				que.push(u);
			}
		}
	}

	for(int u : used)cout << u << endl;
}