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

const int MAX = 1e5;

vector<pair<int, int>> G[MAX];

int a[MAX];

void dfs(int v, int prev) {
	for (auto e : G[v]) if (e.first != prev) {
		a[e.first] = a[v] ^ (e.second & 1);
		dfs(e.first, v);
	}
}

int main()
{
	int N;
	cin >> N;
	for (int i = 0; i < N - 1; i++) {
		int u, v, w;
		cin >> u >> v >> w; --u, --v;
		G[u].emplace_back(v, w & 1);
		G[v].emplace_back(u, w & 1);
	}
	a[0] = 0;
	dfs(0, -1);
	for (int i = 0; i < N; i++) {
		puts(a[i] ? "1" : "0");
	}
	return 0;
}

### Prompt

In cpp, your task is to solve 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 MAX = 1e5;

vector<pair<int, int>> G[MAX];

int a[MAX];

void dfs(int v, int prev) {
	for (auto e : G[v]) if (e.first != prev) {
		a[e.first] = a[v] ^ (e.second & 1);
		dfs(e.first, v);
	}
}

int main()
{
	int N;
	cin >> N;
	for (int i = 0; i < N - 1; i++) {
		int u, v, w;
		cin >> u >> v >> w; --u, --v;
		G[u].emplace_back(v, w & 1);
		G[v].emplace_back(u, w & 1);
	}
	a[0] = 0;
	dfs(0, -1);
	for (int i = 0; i < N; i++) {
		puts(a[i] ? "1" : "0");
	}
	return 0;
}

```