#include <iostream>
#include <vector>

using namespace std;

vector<vector<pair<int, int>>> a(100001);
vector<int> ans(100001);

void dfs(int n, int p, long long t);

int main()
{
	int n;
	cin >> n;
	
	int u, v, w;
	for (int i = 0; i < n - 1; i++) {
		cin >> u >> v >> w;
		u--;
		v--;
		a[u].push_back(make_pair(v, w));
		a[v].push_back(make_pair(u, w));
	}
	
	dfs(0, 0, 0);

	for (int i = 0; i < n; i++) {
		cout << (ans[i] % 2 == 0 ? "0" : "1") << endl;
	}

	return 0;
}

void dfs(int n, int p, long long t)
{
	ans[n] = t;
	for (auto &u : a[n]) {
		if (u.first != p) {
			dfs(u.first, n, t + u.second);
		}
	}
}
### Prompt

Create a solution in CPP for 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;

vector<vector<pair<int, int>>> a(100001);
vector<int> ans(100001);

void dfs(int n, int p, long long t);

int main()
{
	int n;
	cin >> n;
	
	int u, v, w;
	for (int i = 0; i < n - 1; i++) {
		cin >> u >> v >> w;
		u--;
		v--;
		a[u].push_back(make_pair(v, w));
		a[v].push_back(make_pair(u, w));
	}
	
	dfs(0, 0, 0);

	for (int i = 0; i < n; i++) {
		cout << (ans[i] % 2 == 0 ? "0" : "1") << endl;
	}

	return 0;
}

void dfs(int n, int p, long long t)
{
	ans[n] = t;
	for (auto &u : a[n]) {
		if (u.first != p) {
			dfs(u.first, n, t + u.second);
		}
	}
}
```