#include <bits/stdc++.h>

using namespace std;

int n;
int col[100010];
vector<pair<int, int> > adj[100010];

void dfs(int u, int p, bool par){
	if(par) col[u] = 1;
	else col[u] = 0;

	for(auto v : adj[u]){
		if(v.first == p) continue;

		bool tmp = v.second & 1;

		dfs(v.first, u, tmp ^ par);
	}
}

int main(){
	scanf("%d", &n);

	for(int i = 0; i < n - 1; i++){
		int u, v, w;
		scanf("%d%d%d", &u, &v, &w);

		adj[u].emplace_back(v, w);
		adj[v].emplace_back(u, w);
	}

	dfs(1, -1, 0);

	for(int i = 1; i <= n; i++){
		printf("%d\n", col[i]);
	}
	
	return 0;
}

### 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>

using namespace std;

int n;
int col[100010];
vector<pair<int, int> > adj[100010];

void dfs(int u, int p, bool par){
	if(par) col[u] = 1;
	else col[u] = 0;

	for(auto v : adj[u]){
		if(v.first == p) continue;

		bool tmp = v.second & 1;

		dfs(v.first, u, tmp ^ par);
	}
}

int main(){
	scanf("%d", &n);

	for(int i = 0; i < n - 1; i++){
		int u, v, w;
		scanf("%d%d%d", &u, &v, &w);

		adj[u].emplace_back(v, w);
		adj[v].emplace_back(u, w);
	}

	dfs(1, -1, 0);

	for(int i = 1; i <= n; i++){
		printf("%d\n", col[i]);
	}
	
	return 0;
}

```