#include<bits/stdc++.h>
//#include<iostream>
 
using namespace std;
 
typedef long long ll;
 
const int N = 200005;
 
int n;
vector<pair<int,int> > adj[N];

ll dst[N];

void dfs(int x,int p){
	for(auto t:adj[x]){
		int y = t.first,v = t.second;
		if(y == p)continue;
		dst[y] = dst[x] + v;
		dfs(y,x);
	}
}

int main(){
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	
	cin>>n;
	for(int i=1;i<n;i++){
		int x,y,z;
		cin>>x>>y>>z;
		adj[x].push_back({y,z});
		adj[y].push_back({x,z});
	}
	
	dst[1] = 0;
	
	dfs(1,1);
	
	for(int i=1;i<=n;i++){
		cout<<(dst[i]%2)<<endl;
	}
	
	return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
//#include<iostream>
 
using namespace std;
 
typedef long long ll;
 
const int N = 200005;
 
int n;
vector<pair<int,int> > adj[N];

ll dst[N];

void dfs(int x,int p){
	for(auto t:adj[x]){
		int y = t.first,v = t.second;
		if(y == p)continue;
		dst[y] = dst[x] + v;
		dfs(y,x);
	}
}

int main(){
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	
	cin>>n;
	for(int i=1;i<n;i++){
		int x,y,z;
		cin>>x>>y>>z;
		adj[x].push_back({y,z});
		adj[y].push_back({x,z});
	}
	
	dst[1] = 0;
	
	dfs(1,1);
	
	for(int i=1;i<=n;i++){
		cout<<(dst[i]%2)<<endl;
	}
	
	return 0;
}

```