#include<bits/stdc++.h>
using namespace std;
int n;
int fir[100010],nxt[200010],to[200010];
long long w[200010],z;
int cnt;
int x,y;
int a[100010];

void dfs(int x,int f,long long vi){
	if(vi%2==0) a[x]=0;
	else a[x]=1;
	for(int i=fir[x];i;i=nxt[i]){
		if(to[i]==f) continue;
		dfs(to[i],x,vi+w[i]);
	}
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<n;i++){
		scanf("%d %d %lld",&x,&y,&z);
		to[++cnt]=y,nxt[cnt]=fir[x],fir[x]=cnt,w[cnt]=z;
		to[++cnt]=x,nxt[cnt]=fir[y],fir[y]=cnt,w[cnt]=z;
	}
	dfs(1,0,0);
	for(int i=1;i<=n;i++){
		printf("%d\n",a[i]);
	}
	
	return 0;
}
### 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;
int fir[100010],nxt[200010],to[200010];
long long w[200010],z;
int cnt;
int x,y;
int a[100010];

void dfs(int x,int f,long long vi){
	if(vi%2==0) a[x]=0;
	else a[x]=1;
	for(int i=fir[x];i;i=nxt[i]){
		if(to[i]==f) continue;
		dfs(to[i],x,vi+w[i]);
	}
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<n;i++){
		scanf("%d %d %lld",&x,&y,&z);
		to[++cnt]=y,nxt[cnt]=fir[x],fir[x]=cnt,w[cnt]=z;
		to[++cnt]=x,nxt[cnt]=fir[y],fir[y]=cnt,w[cnt]=z;
	}
	dfs(1,0,0);
	for(int i=1;i<=n;i++){
		printf("%d\n",a[i]);
	}
	
	return 0;
}
```