#include<bits/stdc++.h>

namespace nyg
{
	const int N=101000;

	int begin[N],next[N*2],to[N*2],w[N*2];

	int n,e;

	void add(int x,int y,int z,bool k=1)
	{
		to[++e]=y;
		next[e]=begin[x];
		begin[x]=e;
		w[e]=z;
		if(k)add(y,x,z,0);
	}

	bool c[N];
	void dfs(int p=1,int h=0)
	{
		for(int i=begin[p],q;i;i=next[i])
			if((q=to[i])!=h)
				c[q]=c[p]^(w[i]&1),dfs(q,p);
	}

	void solve()
	{
		scanf("%d",&n);
		for(int i=1,u,v,h;i<n;i++)
			scanf("%d%d%d",&u,&v,&h),add(u,v,h);
		dfs();
		for(int i=1;i<=n;i++)
			printf("%d\n",c[i]);
	}
}

int main()
{
	nyg::solve();
	return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>

namespace nyg
{
	const int N=101000;

	int begin[N],next[N*2],to[N*2],w[N*2];

	int n,e;

	void add(int x,int y,int z,bool k=1)
	{
		to[++e]=y;
		next[e]=begin[x];
		begin[x]=e;
		w[e]=z;
		if(k)add(y,x,z,0);
	}

	bool c[N];
	void dfs(int p=1,int h=0)
	{
		for(int i=begin[p],q;i;i=next[i])
			if((q=to[i])!=h)
				c[q]=c[p]^(w[i]&1),dfs(q,p);
	}

	void solve()
	{
		scanf("%d",&n);
		for(int i=1,u,v,h;i<n;i++)
			scanf("%d%d%d",&u,&v,&h),add(u,v,h);
		dfs();
		for(int i=1;i<=n;i++)
			printf("%d\n",c[i]);
	}
}

int main()
{
	nyg::solve();
	return 0;
}

```