#include <bits/stdc++.h>

int ri() {
	int n;
	scanf("%d", &n);
	return n;
}

std::vector<std::vector<std::pair<int, int> > > hen;
std::vector<bool> res;
void dfs(int i, int prev, int64_t dist = 0) {
	res[i] = dist & 1;
	for (auto j : hen[i]) if (j.first != prev) dfs(j.first, i, dist + j.second);
}

int main() {
	int n = ri();
	hen.resize(n);
	for (int i = 1; i < n; i++) {
		int a = ri() - 1;
		int b = ri() - 1;
		int c = ri();
		hen[a].push_back({b, c});
		hen[b].push_back({a, c});
	}
	res.resize(n);
	dfs(0, -1);
	for (auto i : res) printf("%d\n", (int) i);
	
	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>

int ri() {
	int n;
	scanf("%d", &n);
	return n;
}

std::vector<std::vector<std::pair<int, int> > > hen;
std::vector<bool> res;
void dfs(int i, int prev, int64_t dist = 0) {
	res[i] = dist & 1;
	for (auto j : hen[i]) if (j.first != prev) dfs(j.first, i, dist + j.second);
}

int main() {
	int n = ri();
	hen.resize(n);
	for (int i = 1; i < n; i++) {
		int a = ri() - 1;
		int b = ri() - 1;
		int c = ri();
		hen[a].push_back({b, c});
		hen[b].push_back({a, c});
	}
	res.resize(n);
	dfs(0, -1);
	for (auto i : res) printf("%d\n", (int) i);
	
	return 0;
}

```