#include<cstdio>
#define maxn 100050
using namespace std;
struct enode{
	int y,nxt,z;
}e[maxn*2];
int tot=0;
int dis[maxn],first[maxn];
void adde(int x,int y,int z){
	e[tot].nxt=first[x];
	e[tot].y=y;
	e[tot].z=z;
	first[x]=tot++;
}
void dfs(int x,int fa){
	for (int i=first[x];i>=0;i=e[i].nxt){
		int y=e[i].y;
		if (y!=fa){
			dis[y]=(dis[x]+e[i].z)%2;
			dfs(y,x);
		}
	}
}
int main(){
	int n;
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	first[i]=-1;
	for (int i=1;i<n;i++){
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		adde(x,y,z),adde(y,x,z);
	}
	dfs(1,0);
	for (int i=1;i<=n;i++)
	printf("%d\n",dis[i]);
	return 0;
}
### Prompt

Generate 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<cstdio>
#define maxn 100050
using namespace std;
struct enode{
	int y,nxt,z;
}e[maxn*2];
int tot=0;
int dis[maxn],first[maxn];
void adde(int x,int y,int z){
	e[tot].nxt=first[x];
	e[tot].y=y;
	e[tot].z=z;
	first[x]=tot++;
}
void dfs(int x,int fa){
	for (int i=first[x];i>=0;i=e[i].nxt){
		int y=e[i].y;
		if (y!=fa){
			dis[y]=(dis[x]+e[i].z)%2;
			dfs(y,x);
		}
	}
}
int main(){
	int n;
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	first[i]=-1;
	for (int i=1;i<n;i++){
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		adde(x,y,z),adde(y,x,z);
	}
	dfs(1,0);
	for (int i=1;i<=n;i++)
	printf("%d\n",dis[i]);
	return 0;
}
```