#include<bits/stdc++.h>
using namespace std;
vector<pair<int,int> > g[100005];
int ans[100005];
void dfs(int x,int col)
{
	ans[x]=col;
	for(int i=0;i<g[x].size();i++)
		if(g[x][i].second%2==0&&ans[g[x][i].first]==-1)
			dfs(g[x][i].first,col);
		else if(g[x][i].second%2==1&&ans[g[x][i].first]==-1)
			dfs(g[x][i].first,abs(col-1));
}
int main()
{
	int n;
	cin>>n;
	for(int i=1;i<=n-1;i++)
	{
		int u,v,w;
		cin>>u>>v>>w;
		g[u].push_back(make_pair(v,w));
		g[v].push_back(make_pair(u,w));
		ans[i]=-1;
	}
	ans[n]=-1;
	dfs(1,0);
	for(int i=1;i<=n;i++)
		cout<<ans[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<bits/stdc++.h>
using namespace std;
vector<pair<int,int> > g[100005];
int ans[100005];
void dfs(int x,int col)
{
	ans[x]=col;
	for(int i=0;i<g[x].size();i++)
		if(g[x][i].second%2==0&&ans[g[x][i].first]==-1)
			dfs(g[x][i].first,col);
		else if(g[x][i].second%2==1&&ans[g[x][i].first]==-1)
			dfs(g[x][i].first,abs(col-1));
}
int main()
{
	int n;
	cin>>n;
	for(int i=1;i<=n-1;i++)
	{
		int u,v,w;
		cin>>u>>v>>w;
		g[u].push_back(make_pair(v,w));
		g[v].push_back(make_pair(u,w));
		ans[i]=-1;
	}
	ans[n]=-1;
	dfs(1,0);
	for(int i=1;i<=n;i++)
		cout<<ans[i]<<endl;
	return 0;
}

```