#include <cstdio>
const int maxn = 1e5+5;
int n,col[maxn];
int head[maxn],tot;
struct Edge{
	int next,to,val;
}e[maxn<<1];
inline void add_edge(int x,int y,int z){
	e[++tot].next = head[x];
	e[tot].to = y,e[tot].val = z,head[x] = tot;
}
inline void dfs(int u,int F,int c){
	col[u] = c;
	for(int i=head[u],v;v=e[i].to,i;i=e[i].next)
		if(v != F){
			if(e[i].val&1)dfs(v,u,c^1);
			else dfs(v,u,c);
		}
}
int main(){
	scanf("%d",&n);
	for(int i=1,x,y,z;i<n;++i){
		scanf("%d %d %d",&x,&y,&z);
		add_edge(x,y,z),add_edge(y,x,z);
	}
	dfs(1,0,1);
	for(int i=1;i<=n;++i)printf("%d\n",col[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 <cstdio>
const int maxn = 1e5+5;
int n,col[maxn];
int head[maxn],tot;
struct Edge{
	int next,to,val;
}e[maxn<<1];
inline void add_edge(int x,int y,int z){
	e[++tot].next = head[x];
	e[tot].to = y,e[tot].val = z,head[x] = tot;
}
inline void dfs(int u,int F,int c){
	col[u] = c;
	for(int i=head[u],v;v=e[i].to,i;i=e[i].next)
		if(v != F){
			if(e[i].val&1)dfs(v,u,c^1);
			else dfs(v,u,c);
		}
}
int main(){
	scanf("%d",&n);
	for(int i=1,x,y,z;i<n;++i){
		scanf("%d %d %d",&x,&y,&z);
		add_edge(x,y,z),add_edge(y,x,z);
	}
	dfs(1,0,1);
	for(int i=1;i<=n;++i)printf("%d\n",col[i]);
	return 0;
}
```