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

vector<pair<int,int>> g[100002];
int ans[100002];

void dfs(int x,int p,int w){
	ans[x] = w;
	//cout<<"dfdf";
	for(auto a: g[x]){
		if(a.first!=p){
			dfs(a.first,x,w^a.second);
		}
	}
}

int main(){
	int n;
	cin>>n;

	for(int i = 0; i < n-1; i++){
		int x,y,w;
		cin>>x>>y>>w;
		g[x].push_back({y,w%2});
		g[y].push_back({x,w%2});
	}

	dfs(1,0,0);
	for(int i = 1; i<=n; i++){
		cout<<ans[i]<<"\n";
	}

}
### Prompt

Generate 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;

vector<pair<int,int>> g[100002];
int ans[100002];

void dfs(int x,int p,int w){
	ans[x] = w;
	//cout<<"dfdf";
	for(auto a: g[x]){
		if(a.first!=p){
			dfs(a.first,x,w^a.second);
		}
	}
}

int main(){
	int n;
	cin>>n;

	for(int i = 0; i < n-1; i++){
		int x,y,w;
		cin>>x>>y>>w;
		g[x].push_back({y,w%2});
		g[y].push_back({x,w%2});
	}

	dfs(1,0,0);
	for(int i = 1; i<=n; i++){
		cout<<ans[i]<<"\n";
	}

}
```