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:
// In the name of God

#include <bits/stdc++.h>

using namespace std;

const int N = 1e5+5;

int n, c[N];
vector<pair<int,int>> adj[N];

void dfs(int u, int p, int col) {
	c[u] = col;
	for(auto v : adj[u]) {
		if(v.first == p) continue;
		dfs(v.first, u, col ^ v.second);
	}
}	

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	cin >> n;

	for(int i = 0; i < n - 1; i++) {
		int u, v, w;
		cin >> u >> v >> w;
		w = w & 1;
		u--;
		v--;
		adj[u].push_back(make_pair(v, w));
		adj[v].push_back(make_pair(u, w));
	}

	dfs(0, 0, 0);

	for(int i = 0; i < n; i++) cout << c[i] << "\n";

	return 0;
}