#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
 
const int maxn = 1e5 + 10;

vector<pair<int, int> > t[maxn];
int c[maxn];

void dfs(int v, int par = -1){
	for (auto it : t[v]){
		int u = it.first, w = it.second % 2;
		if (u != par){
			c[u] = c[v] ^ w;
			dfs(u, v);
		}
	}
}

int main(){
	ios_base::sync_with_stdio (false);
	int n;
	cin >> n;
	for (int i = 0; i < n - 1; i++){
		int v, u, w;
		cin >> v >> u >> w;
		t[v].push_back({u, w});
		t[u].push_back({v, w});
	}
	dfs(1);
	for (int i = 1; i <= n; i++)
		cout << c[i] << '\n';
}

### Prompt

Your challenge is to write 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>

using namespace std;

typedef long long ll;
 
const int maxn = 1e5 + 10;

vector<pair<int, int> > t[maxn];
int c[maxn];

void dfs(int v, int par = -1){
	for (auto it : t[v]){
		int u = it.first, w = it.second % 2;
		if (u != par){
			c[u] = c[v] ^ w;
			dfs(u, v);
		}
	}
}

int main(){
	ios_base::sync_with_stdio (false);
	int n;
	cin >> n;
	for (int i = 0; i < n - 1; i++){
		int v, u, w;
		cin >> v >> u >> w;
		t[v].push_back({u, w});
		t[u].push_back({v, w});
	}
	dfs(1);
	for (int i = 1; i <= n; i++)
		cout << c[i] << '\n';
}

```