#include<bits/stdc++.h>
using namespace std;

const int N=1e5+5;
const int M=3e5+5;
int n,u,v,x,ans[N];
int E,ot[M],ne[M],g[N],w[M];

void add(int u,int v,int x)
{
	ot[++E]=v;
	ne[E]=g[u];
	g[u]=E;
	w[E]=x&1;
}

void init()
{
	scanf("%d",&n);int i;
	for(i=1;i<n;++i)
	{
		scanf("%d%d%d",&u,&v,&x);
		add(u,v,x),add(v,u,x);
	}
}

void dfs(int x,int fath)
{
	int p,to;
	for(p=g[x];p;p=ne[p])
	{
		to=ot[p];
		if(to==fath) continue;
		if(!w[p]) ans[to]=ans[x];
		else ans[to]=ans[x]^1;
		dfs(to,x);
	}
}

int main()
{
	init();dfs(1,0);int i;
	for(i=1;i<=n;++i) printf("%d\n",ans[i]);
	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>
using namespace std;

const int N=1e5+5;
const int M=3e5+5;
int n,u,v,x,ans[N];
int E,ot[M],ne[M],g[N],w[M];

void add(int u,int v,int x)
{
	ot[++E]=v;
	ne[E]=g[u];
	g[u]=E;
	w[E]=x&1;
}

void init()
{
	scanf("%d",&n);int i;
	for(i=1;i<n;++i)
	{
		scanf("%d%d%d",&u,&v,&x);
		add(u,v,x),add(v,u,x);
	}
}

void dfs(int x,int fath)
{
	int p,to;
	for(p=g[x];p;p=ne[p])
	{
		to=ot[p];
		if(to==fath) continue;
		if(!w[p]) ans[to]=ans[x];
		else ans[to]=ans[x]^1;
		dfs(to,x);
	}
}

int main()
{
	init();dfs(1,0);int i;
	for(i=1;i<=n;++i) printf("%d\n",ans[i]);
	return 0;
}
```