#include <iostream>
#include <vector>
using namespace std;
 
const int MAXN = 1e5 + 10;
vector<pair<int, int> > Mat[MAXN];
bool cl[MAXN];
int n;
 
void dfs(int v, int p, int s) {
	cl[v] = s % 2;
	for (int i = 0; i < Mat[v].size(); i++)
		if (Mat[v][i].first != p)
			dfs(Mat[v][i].first, v, s + Mat[v][i].second);
}
 
int main() {
	cin >> n;
	for (int i = 0; i < n - 1; i++) {
		int a, b, w;
		cin >> a >> b >> w;
		Mat[a].push_back({b, w % 2});
		Mat[b].push_back({a, w % 2});
	}
	dfs(1, 1, 0);
	for (int i = 1; i <= n; i++)
		cout << cl[i] << '\n';
}
### Prompt

Your task is to create a Cpp solution to 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 <iostream>
#include <vector>
using namespace std;
 
const int MAXN = 1e5 + 10;
vector<pair<int, int> > Mat[MAXN];
bool cl[MAXN];
int n;
 
void dfs(int v, int p, int s) {
	cl[v] = s % 2;
	for (int i = 0; i < Mat[v].size(); i++)
		if (Mat[v][i].first != p)
			dfs(Mat[v][i].first, v, s + Mat[v][i].second);
}
 
int main() {
	cin >> n;
	for (int i = 0; i < n - 1; i++) {
		int a, b, w;
		cin >> a >> b >> w;
		Mat[a].push_back({b, w % 2});
		Mat[b].push_back({a, w % 2});
	}
	dfs(1, 1, 0);
	for (int i = 1; i <= n; i++)
		cout << cl[i] << '\n';
}
```