#include <bits/stdc++.h>
#define N 100005
using namespace std;
int n, m, i, j, k, a, b, c, h[N], v[N];
struct AB{
	int a, b, c, n;
} d[N*2];
void cun(int a, int b, int c){
	d[++k].n = h[a], h[a] = k;
	d[k].a = a, d[k].b = b, d[k].c = c;
}
void dfs(int a){
	int b, c, i;
	for(i=h[a]; i; i=d[i].n){
		b = d[i].b, c = d[i].c;
		if(!v[b]){
			v[b] = v[a] + c;
			dfs(b);
		}
	}
}
int main(){
	scanf("%d", &n);
	for(i=1; i<n; i++){
		scanf("%d%d%d", &a, &b, &c);
		cun(a, b, c);
		cun(b, a, c);
	}
	dfs(v[1] = 1);
	for(i=1; i<=n; i++){
		printf("%d\n", v[i]&1);
	}
	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 <bits/stdc++.h>
#define N 100005
using namespace std;
int n, m, i, j, k, a, b, c, h[N], v[N];
struct AB{
	int a, b, c, n;
} d[N*2];
void cun(int a, int b, int c){
	d[++k].n = h[a], h[a] = k;
	d[k].a = a, d[k].b = b, d[k].c = c;
}
void dfs(int a){
	int b, c, i;
	for(i=h[a]; i; i=d[i].n){
		b = d[i].b, c = d[i].c;
		if(!v[b]){
			v[b] = v[a] + c;
			dfs(b);
		}
	}
}
int main(){
	scanf("%d", &n);
	for(i=1; i<n; i++){
		scanf("%d%d%d", &a, &b, &c);
		cun(a, b, c);
		cun(b, a, c);
	}
	dfs(v[1] = 1);
	for(i=1; i<=n; i++){
		printf("%d\n", v[i]&1);
	}
	return 0;
}
```