#include <iostream>
#include <vector>
using namespace std;
struct edge{int to,cost;};
vector<edge> G[100010];
int d[100010];
void dfs(int s,int p,int c){
	for(auto v:G[s]){
		if(v.to==p) continue;
		if(v.cost&1){
			d[v.to] = 1 - c;
			dfs(v.to,s,1 - c);
		}else{
			d[v.to] = c;
			dfs(v.to,s,c);
		}
	}
}
int main(){
	int i,n;
	cin >> n;
	for(i=0;i<n-1;i++){
		int a,b,c;
		cin >> a >> b >> c;
		a--; b--;
		G[a].push_back({b,c});
		G[b].push_back({a,c});
	}
	d[0] = 0;
	dfs(0,-1,0);
	for(i=0;i<n;i++){
		cout << d[i] << endl;
	}
}
	
### Prompt

Your challenge is to write 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 <iostream>
#include <vector>
using namespace std;
struct edge{int to,cost;};
vector<edge> G[100010];
int d[100010];
void dfs(int s,int p,int c){
	for(auto v:G[s]){
		if(v.to==p) continue;
		if(v.cost&1){
			d[v.to] = 1 - c;
			dfs(v.to,s,1 - c);
		}else{
			d[v.to] = c;
			dfs(v.to,s,c);
		}
	}
}
int main(){
	int i,n;
	cin >> n;
	for(i=0;i<n-1;i++){
		int a,b,c;
		cin >> a >> b >> c;
		a--; b--;
		G[a].push_back({b,c});
		G[b].push_back({a,c});
	}
	d[0] = 0;
	dfs(0,-1,0);
	for(i=0;i<n;i++){
		cout << d[i] << endl;
	}
}
	
```