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

#define int long long
typedef pair <int, int> pii;

const int N = 1e5 + 10;
int n, ans[N];
vector <pii> adj[N];

void dfs(int v, int p, int d) {
	ans[v] = (d & 1);
	for (pii i : adj[v])
		if (i.first != p)
			dfs(i.first, v, d + i.second);
}

int32_t main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n;
	for (int i = 1; i < n; i++) {
		int v, u, w;
		cin >> v >> u >> w;
		v--; u--;
		adj[v].push_back({u, w});
		adj[u].push_back({v, w});
	}
	dfs(0, 0, 0);
	
	for (int i = 0; i < n; i++)
		cout << ans[i] << '\n';
}
### 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;

#define int long long
typedef pair <int, int> pii;

const int N = 1e5 + 10;
int n, ans[N];
vector <pii> adj[N];

void dfs(int v, int p, int d) {
	ans[v] = (d & 1);
	for (pii i : adj[v])
		if (i.first != p)
			dfs(i.first, v, d + i.second);
}

int32_t main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n;
	for (int i = 1; i < n; i++) {
		int v, u, w;
		cin >> v >> u >> w;
		v--; u--;
		adj[v].push_back({u, w});
		adj[u].push_back({v, w});
	}
	dfs(0, 0, 0);
	
	for (int i = 0; i < n; i++)
		cout << ans[i] << '\n';
}
```