#include<bits/stdc++.h>
using namespace std;
int n;
struct Node{
	int id,v;
};
int ans[100010];
vector<Node> q[100010];
void dfs(int id,int fa,int tot)
{
	if(tot%2==0)
		ans[id]=0;
	else
		ans[id]=1;
	for(int i=0;i<q[id].size();i++)
	{
		if(q[id][i].id!=fa)
		dfs(q[id][i].id,id,q[id][i].v+tot);
	}
}
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	cin>>n;
	for(int i=1;i<n;i++)
	{
		int x,y,w;
		Node aa;
		cin>>x>>y>>w;
		aa.id=y,
		aa.v=w;
		q[x].push_back(aa);
		aa.id=x,
		aa.v=w;
		q[y].push_back(aa);
	}
	dfs(1,0,0);
	for(int i=1;i<=n;i++)
		cout<<ans[i]<<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>
using namespace std;
int n;
struct Node{
	int id,v;
};
int ans[100010];
vector<Node> q[100010];
void dfs(int id,int fa,int tot)
{
	if(tot%2==0)
		ans[id]=0;
	else
		ans[id]=1;
	for(int i=0;i<q[id].size();i++)
	{
		if(q[id][i].id!=fa)
		dfs(q[id][i].id,id,q[id][i].v+tot);
	}
}
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	cin>>n;
	for(int i=1;i<n;i++)
	{
		int x,y,w;
		Node aa;
		cin>>x>>y>>w;
		aa.id=y,
		aa.v=w;
		q[x].push_back(aa);
		aa.id=x,
		aa.v=w;
		q[y].push_back(aa);
	}
	dfs(1,0,0);
	for(int i=1;i<=n;i++)
		cout<<ans[i]<<endl;
	return 0;
}
```