#include<bits/stdc++.h>
using namespace std;
int n;
struct edge{
	int v,w,next;
}e[405000];
int head[405000],cnt;
int ans[405000];
void add(int u,int v,int w){
	++cnt;
	e[cnt].v=v;
	e[cnt].w=w;
	e[cnt].next=head[u];
	head[u]=cnt;
}
void dfs(int u,int fa,int sum){
	ans[u]=sum;
	for(int i=head[u];i;i=e[i].next){
		int v=e[i].v;
		if(v==fa)continue;
		if(e[i].w%2)dfs(v,u,sum^1);
		else dfs(v,u,sum);
	}
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n-1;++i){
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w);
		add(u,v,w);add(v,u,w);
	}
	dfs(1,-1,0);
	for(int i=1;i<=n;++i)cout<<ans[i]<<endl; 
}
### Prompt

Create a solution in cpp for 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;
int n;
struct edge{
	int v,w,next;
}e[405000];
int head[405000],cnt;
int ans[405000];
void add(int u,int v,int w){
	++cnt;
	e[cnt].v=v;
	e[cnt].w=w;
	e[cnt].next=head[u];
	head[u]=cnt;
}
void dfs(int u,int fa,int sum){
	ans[u]=sum;
	for(int i=head[u];i;i=e[i].next){
		int v=e[i].v;
		if(v==fa)continue;
		if(e[i].w%2)dfs(v,u,sum^1);
		else dfs(v,u,sum);
	}
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n-1;++i){
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w);
		add(u,v,w);add(v,u,w);
	}
	dfs(1,-1,0);
	for(int i=1;i<=n;++i)cout<<ans[i]<<endl; 
}
```