#include <bits/stdc++.h>
using namespace std;
const int n = 1e5+5;
int N;
vector<pair<int, bool>> G[n];
bool vis[n];
bool col[n];

void DFS(int v, int pr, bool c){
	col[v] = c;
	vis[v] = true;
	for(auto p : G[v]){
		int j = p.first;
		bool cg = p.second;
		if(!vis[j] && j != pr){
			DFS(j, v, c^cg);
		}
	}
}

int main(){
	cin >> N;
	for(int i = 0; i < N-1; ++i){
		int u,v,w;
		cin >> u >> v >> w;
		u--;v--;
		G[u].push_back({v, w%2});
		G[v].push_back({u, w%2});
	}
	for(int i =0; i < N; ++i){
		if(!vis[i]){
			DFS(i, -1, true);
		}
		cout << col[i] << endl;
	}
	return 0;
}
### Prompt

Generate 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 <bits/stdc++.h>
using namespace std;
const int n = 1e5+5;
int N;
vector<pair<int, bool>> G[n];
bool vis[n];
bool col[n];

void DFS(int v, int pr, bool c){
	col[v] = c;
	vis[v] = true;
	for(auto p : G[v]){
		int j = p.first;
		bool cg = p.second;
		if(!vis[j] && j != pr){
			DFS(j, v, c^cg);
		}
	}
}

int main(){
	cin >> N;
	for(int i = 0; i < N-1; ++i){
		int u,v,w;
		cin >> u >> v >> w;
		u--;v--;
		G[u].push_back({v, w%2});
		G[v].push_back({u, w%2});
	}
	for(int i =0; i < N; ++i){
		if(!vis[i]){
			DFS(i, -1, true);
		}
		cout << col[i] << endl;
	}
	return 0;
}
```