#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int>> a[100005];
int ans[100005];
void dfs(int no,int father){ 
	for (int i=0;i<a[no].size();i++){
		int u=a[no][i].first;
		int w=a[no][i].second%2; 
		if(u!=father){
         ans[u]=ans[no]^w;
         dfs(u,no);
       }
  }
}
int main(){
    int n;
	cin>>n;
	for (int i=1;i<n;i++){
		int u,v,c;
		cin>>u>>v>>c;
		a[u].push_back({v,c});
		a[v].push_back({u,c});
	}
	dfs(1,-1);
	for(int i=1;i<=n;i++) cout<<ans[i]<<endl;
}

### Prompt

Please provide a cpp coded solution 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;
vector<pair<int, int>> a[100005];
int ans[100005];
void dfs(int no,int father){ 
	for (int i=0;i<a[no].size();i++){
		int u=a[no][i].first;
		int w=a[no][i].second%2; 
		if(u!=father){
         ans[u]=ans[no]^w;
         dfs(u,no);
       }
  }
}
int main(){
    int n;
	cin>>n;
	for (int i=1;i<n;i++){
		int u,v,c;
		cin>>u>>v>>c;
		a[u].push_back({v,c});
		a[v].push_back({u,c});
	}
	dfs(1,-1);
	for(int i=1;i<=n;i++) cout<<ans[i]<<endl;
}

```