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

#define LL long long

int color[100009];

vector<pair<int,int> > G[100009];

void dfs(int s,int p,int dis){
	color[s] = dis;
	for(int i=0;i<(int)G[s].size();i++){
		pair<int,int> p1 = G[s][i];
		if(p1.first != p){
			dfs(p1.first,s,(dis+p1.second)%2);
		}
	}
}

int main(){
	

	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	int n;
	cin>>n;

	for(int i=1;i<n;i++){
		int a,b,w;
		cin>>a>>b>>w;

		G[a].push_back(make_pair(b,w%2));
		G[b].push_back(make_pair(a,w%2));
	}

	dfs(1,1,0);

	for(int i=1;i<=n;i++){
		cout<<color[i]<<endl;
	}
}
### Prompt

Your task is to create 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;

#define LL long long

int color[100009];

vector<pair<int,int> > G[100009];

void dfs(int s,int p,int dis){
	color[s] = dis;
	for(int i=0;i<(int)G[s].size();i++){
		pair<int,int> p1 = G[s][i];
		if(p1.first != p){
			dfs(p1.first,s,(dis+p1.second)%2);
		}
	}
}

int main(){
	

	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	int n;
	cin>>n;

	for(int i=1;i<n;i++){
		int a,b,w;
		cin>>a>>b>>w;

		G[a].push_back(make_pair(b,w%2));
		G[b].push_back(make_pair(a,w%2));
	}

	dfs(1,1,0);

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