#include <bits/stdc++.h>

using namespace std;

vector<pair<int, int> > G[101010];
int ans[101010];
void dfs(int a, int ta, int c) {
	ans[a] = c;
	for (auto x : G[a]) {
		if (x.first != ta) 
			dfs(x.first, a, x.second ^ c);
	}
}


int main() {
	int V;
	cin >> V;
	for (int i = 0; i < V - 1; i++) {
		int s, t, c;
		cin >> s >> t >> c;
		G[s].emplace_back(t, c % 2);// c even -> 0
		G[t].emplace_back(s, c % 2); 
	}
	dfs(1, -1, 0);
	for (int i = 1; i <= V; i++)
		cout << ans[i] <<endl;
	return 0;
}


### Prompt

Create a solution in cpp for the following 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

### Response

```cpp
#include <bits/stdc++.h>

using namespace std;

vector<pair<int, int> > G[101010];
int ans[101010];
void dfs(int a, int ta, int c) {
	ans[a] = c;
	for (auto x : G[a]) {
		if (x.first != ta) 
			dfs(x.first, a, x.second ^ c);
	}
}


int main() {
	int V;
	cin >> V;
	for (int i = 0; i < V - 1; i++) {
		int s, t, c;
		cin >> s >> t >> c;
		G[s].emplace_back(t, c % 2);// c even -> 0
		G[t].emplace_back(s, c % 2); 
	}
	dfs(1, -1, 0);
	for (int i = 1; i <= V; i++)
		cout << ans[i] <<endl;
	return 0;
}


```