#include<bits/stdc++.h>
using namespace std;
const int maxn=2e5+5;
struct node{
	int to,nxt;
	int w;
}e[maxn];
int head[maxn];
int ans[maxn];
int cnt;
void add(int u,int v,int w){
	e[++cnt]=(node){v,head[u],w};
	head[u]=cnt;
}
void dfs(int u,int fa){
	for(int i=head[u];i;i=e[i].nxt){
		int v=e[i].to;
		if(v==fa)
			continue;
		if(e[i].w%2)
			ans[v]=ans[u]^1;
		else ans[v]=ans[u];
		dfs(v,u);
	}
}
int main(){
	int n;
	scanf("%d",&n);
	for(int i,u,v,w;i<n-1;i++){
		scanf("%d%d%d",&u,&v,&w);
		add(u,v,w);
		add(v,u,w);
	}
	dfs(1,0);
	for(int i=1;i<=n;i++)
		printf("%d\n",ans[i]);
}
### Prompt

Your task is to create a CPP solution to 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;
const int maxn=2e5+5;
struct node{
	int to,nxt;
	int w;
}e[maxn];
int head[maxn];
int ans[maxn];
int cnt;
void add(int u,int v,int w){
	e[++cnt]=(node){v,head[u],w};
	head[u]=cnt;
}
void dfs(int u,int fa){
	for(int i=head[u];i;i=e[i].nxt){
		int v=e[i].to;
		if(v==fa)
			continue;
		if(e[i].w%2)
			ans[v]=ans[u]^1;
		else ans[v]=ans[u];
		dfs(v,u);
	}
}
int main(){
	int n;
	scanf("%d",&n);
	for(int i,u,v,w;i<n-1;i++){
		scanf("%d%d%d",&u,&v,&w);
		add(u,v,w);
		add(v,u,w);
	}
	dfs(1,0);
	for(int i=1;i<=n;i++)
		printf("%d\n",ans[i]);
}
```