#include "bits/stdc++.h"
using namespace std;

long long N;
static vector<pair<int, int> > V[100000];
static long long Dist[100000] = {};
static bool visited[100000] = {};

void DFS(int S) {
	visited[S] = true;
	for (pair<int, int> NS : V[S]) {
		if (!visited[NS.first]) {
			Dist[NS.first] = (Dist[S] + NS.second) % 2;
			DFS(NS.first);
		}
	}
}

int main() {
	cin >> N;
	for (int i = 0; i < N - 1; i++) {
		int u, v, w;
		cin >> u >> v >> w;
		w = w % 2;
		u--, v--;
		V[u].push_back({ v, w });
		V[v].push_back({ u, w });
	}
	DFS(0);
	for (int i = 0; i < N; i++) {
		cout << Dist[i] << endl;
	}
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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;

long long N;
static vector<pair<int, int> > V[100000];
static long long Dist[100000] = {};
static bool visited[100000] = {};

void DFS(int S) {
	visited[S] = true;
	for (pair<int, int> NS : V[S]) {
		if (!visited[NS.first]) {
			Dist[NS.first] = (Dist[S] + NS.second) % 2;
			DFS(NS.first);
		}
	}
}

int main() {
	cin >> N;
	for (int i = 0; i < N - 1; i++) {
		int u, v, w;
		cin >> u >> v >> w;
		w = w % 2;
		u--, v--;
		V[u].push_back({ v, w });
		V[v].push_back({ u, w });
	}
	DFS(0);
	for (int i = 0; i < N; i++) {
		cout << Dist[i] << endl;
	}
}
```