#include <bits/stdc++.h>
#include <ctype.h>

using namespace std;
using LL = long long;

const int N = 1e5 + 5;
vector< pair< int, int>> conn[N];
int color[N];
bool seen[N];

void dfs(int a, int d) {
	seen[a] = true;
	color[a] = d % 2;
	for (auto i : conn[a]) {
		if (seen[i.first]) continue;
		dfs(i.first, d + i.second);
	}
}	

int main() {
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n - 1; i ++) {
		int a, b, c;
		scanf("%d %d %d", &a, &b, &c);
		conn[a].emplace_back(b, c % 2);
		conn[b].emplace_back(a, c % 2);
	}

	dfs(1, 0);
	for (int i = 1; i <= n; ++i) printf("%d\n", color[i]);
}
### Prompt

Please create a solution in CPP 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 <bits/stdc++.h>
#include <ctype.h>

using namespace std;
using LL = long long;

const int N = 1e5 + 5;
vector< pair< int, int>> conn[N];
int color[N];
bool seen[N];

void dfs(int a, int d) {
	seen[a] = true;
	color[a] = d % 2;
	for (auto i : conn[a]) {
		if (seen[i.first]) continue;
		dfs(i.first, d + i.second);
	}
}	

int main() {
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n - 1; i ++) {
		int a, b, c;
		scanf("%d %d %d", &a, &b, &c);
		conn[a].emplace_back(b, c % 2);
		conn[b].emplace_back(a, c % 2);
	}

	dfs(1, 0);
	for (int i = 1; i <= n; ++i) printf("%d\n", color[i]);
}
```