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 N = 2e5 + 10;

int color[N] ;
vector<pair<int ,int > > v[N];


void dfs(int i , int col , int par = -1) {
	
	for(auto j : v[i]) {
		int node = j.first , c = j.second;
		if(node == par) continue;
		if(c % 2 == 0) color[node] = col;
		else color[node] = col^1;
		dfs(node ,color[node] , i);
	}
	
}
		


int main() {
	int n ; cin >> n;
	for(int i = 0 ;i < n - 1; i ++) {
		int x , y, w ; cin >> x >> y >> w;
		v[x].push_back({y , w});
		v[y].push_back({x , w});
	}
	
	dfs(1 , 0);
	for(int i = 1; i <= n ; i ++) cout << color[i] << endl;
}