#include<iostream>
#include<vector>
#include<utility>
using namespace std;
int n;
vector<pair<int,int> >v[100000];
int c[100000];
void dfs(int vv,int p=-1){
	int u,w;
	for(int i=0;i<v[vv].size();++i){
		u=v[vv][i].first;
		w=v[vv][i].second&1;
		if(u!=p){
			c[u]=c[vv]^w;
			dfs(u,vv);
		}
	}
}
int main(){
	cin>>n;
	int u,vv,w;
	for(int i=0;i<n-1;++i){
		cin>>u>>vv>>w;
		v[u-1].push_back(make_pair(vv-1,w));
		v[vv-1].push_back(make_pair(u-1,w));
	}
	dfs(0);
	for(int i=0;i<n;++i)
		cout<<c[i]<<endl;
	return 0;
}
### 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>
#include<utility>
using namespace std;
int n;
vector<pair<int,int> >v[100000];
int c[100000];
void dfs(int vv,int p=-1){
	int u,w;
	for(int i=0;i<v[vv].size();++i){
		u=v[vv][i].first;
		w=v[vv][i].second&1;
		if(u!=p){
			c[u]=c[vv]^w;
			dfs(u,vv);
		}
	}
}
int main(){
	cin>>n;
	int u,vv,w;
	for(int i=0;i<n-1;++i){
		cin>>u>>vv>>w;
		v[u-1].push_back(make_pair(vv-1,w));
		v[vv-1].push_back(make_pair(u-1,w));
	}
	dfs(0);
	for(int i=0;i<n;++i)
		cout<<c[i]<<endl;
	return 0;
}
```