#include<iostream>
#include<vector>
using namespace::std;

vector<pair<int,int>> v[100005];
int ar[100005];

void DFS(int i){
	for(auto a:v[i])
		if(!ar[a.first])
			if(a.second&1) ar[a.first]=(ar[i]&1)+1,DFS(a.first);
			else ar[a.first]=ar[i],DFS(a.first);
}

int main(){
	int n;
	cin>>n;
	for(int i=0,l,r,w;i<n-1;i++)
		cin>>l>>r>>w,v[l].push_back({r,w}),v[r].push_back({l,w});
	ar[1]=1;
	DFS(1);
	for(int i=1;i<=n;i++)
		cout<<ar[i]-1<<'\n';
}
### Prompt

In Cpp, your task is to solve 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<pair<int,int>> v[100005];
int ar[100005];

void DFS(int i){
	for(auto a:v[i])
		if(!ar[a.first])
			if(a.second&1) ar[a.first]=(ar[i]&1)+1,DFS(a.first);
			else ar[a.first]=ar[i],DFS(a.first);
}

int main(){
	int n;
	cin>>n;
	for(int i=0,l,r,w;i<n-1;i++)
		cin>>l>>r>>w,v[l].push_back({r,w}),v[r].push_back({l,w});
	ar[1]=1;
	DFS(1);
	for(int i=1;i<=n;i++)
		cout<<ar[i]-1<<'\n';
}
```