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

int n;
int a,b,c;

int ans[100002];

vector<pair<int,int> >vec[100002];

int slv(int now,int col){
	ans[now]=col;
	for(int i=0;i<vec[now].size();i++){
		int next=vec[now][i].first , dir=vec[now][i].second;
		if(ans[next]!=-1)continue;
		if(dir%2==1){
			slv(next,col^1);
		}
		else{
			slv(next,col);
		}
	}
	return(0);
}

int main(){
	
	cin>>n;
	for(int i=1;i<=n;i++)ans[i]=-1;
	for(int i=1;i<n;i++){
		cin>>a>>b>>c;
		vec[a].push_back(make_pair(b,c));
		vec[b].push_back(make_pair(a,c));
	}

	slv(1,0);

	for(int i=1;i<=n;i++){
		cout<<ans[i]<<endl;
	}
	return(0);
}
### 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;

int n;
int a,b,c;

int ans[100002];

vector<pair<int,int> >vec[100002];

int slv(int now,int col){
	ans[now]=col;
	for(int i=0;i<vec[now].size();i++){
		int next=vec[now][i].first , dir=vec[now][i].second;
		if(ans[next]!=-1)continue;
		if(dir%2==1){
			slv(next,col^1);
		}
		else{
			slv(next,col);
		}
	}
	return(0);
}

int main(){
	
	cin>>n;
	for(int i=1;i<=n;i++)ans[i]=-1;
	for(int i=1;i<n;i++){
		cin>>a>>b>>c;
		vec[a].push_back(make_pair(b,c));
		vec[b].push_back(make_pair(a,c));
	}

	slv(1,0);

	for(int i=1;i<=n;i++){
		cout<<ans[i]<<endl;
	}
	return(0);
}
```