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

const int MAXN = 100010;
int N;
vector<pair<int, int>> adj[MAXN];
int ans[MAXN];

void dfs(int cur = 1, int prv = 0) {
	for (auto e : adj[cur]) {
		int nxt, w; tie(nxt, w) = e;
		if (nxt == prv) continue;
		ans[nxt] = (ans[cur] + w) % 2;
		dfs(nxt, cur);
	}
}

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> N;
	for (int i = 0; i < N-1; i++) {
		int a, b, c; cin >> a >> b >> c; c %= 2;
		adj[a].emplace_back(b, c);
		adj[b].emplace_back(a, c);
	}
	dfs();
	for (int i = 1; i <= N; i++) {
		cout << ans[i] << '\n';
	}

	return 0;
}

### Prompt

Develop a solution in cpp 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;

const int MAXN = 100010;
int N;
vector<pair<int, int>> adj[MAXN];
int ans[MAXN];

void dfs(int cur = 1, int prv = 0) {
	for (auto e : adj[cur]) {
		int nxt, w; tie(nxt, w) = e;
		if (nxt == prv) continue;
		ans[nxt] = (ans[cur] + w) % 2;
		dfs(nxt, cur);
	}
}

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> N;
	for (int i = 0; i < N-1; i++) {
		int a, b, c; cin >> a >> b >> c; c %= 2;
		adj[a].emplace_back(b, c);
		adj[b].emplace_back(a, c);
	}
	dfs();
	for (int i = 1; i <= N; i++) {
		cout << ans[i] << '\n';
	}

	return 0;
}

```