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

const int N = (int) 1e5 + 2;
vector<pair<int, int>> adj[N];
int col[N];
int n;
void dfs(int x, int fa) {
	for (auto& e : adj[x]) if (e.first != fa) {
		if (e.second % 2 == 1) {
			col[e.first] = col[x] ^ 1;
		} else {
			col[e.first] = col[x];
		}
		dfs(e.first, x);
	}
}
int main() {
	cin.sync_with_stdio(0);
	cin.tie(0);
	cin >> n;
	for (int i = 0; i < n - 1; ++i) {
		int u, v, w;
		cin >> u >> v >> w, --u, --v;
		adj[u].push_back({v, w});
		adj[v].push_back({u, w});
	}
	col[0] = 0;
	dfs(0, -1);
	for (int i = 0; i < n; ++i) cout << col[i] << "\n";
	return 0;
}

### 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 <bits/stdc++.h>
using namespace std;

const int N = (int) 1e5 + 2;
vector<pair<int, int>> adj[N];
int col[N];
int n;
void dfs(int x, int fa) {
	for (auto& e : adj[x]) if (e.first != fa) {
		if (e.second % 2 == 1) {
			col[e.first] = col[x] ^ 1;
		} else {
			col[e.first] = col[x];
		}
		dfs(e.first, x);
	}
}
int main() {
	cin.sync_with_stdio(0);
	cin.tie(0);
	cin >> n;
	for (int i = 0; i < n - 1; ++i) {
		int u, v, w;
		cin >> u >> v >> w, --u, --v;
		adj[u].push_back({v, w});
		adj[v].push_back({u, w});
	}
	col[0] = 0;
	dfs(0, -1);
	for (int i = 0; i < n; ++i) cout << col[i] << "\n";
	return 0;
}

```