// D - Even Relation
#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using NODE = vector<pair<int,int>>;
using TREE = vector<NODE>;

void dfs(TREE &T, vi &dist, int p, int w){
	if(dist[p] >= 0) return;
	dist[p] = w;
	for(auto c:T[p]){
		int np,nw; tie(np,nw) = c;
		dfs(T, dist, np, w^nw);
	}
}

int main(){
	int N; cin>>N;
	vi dist(N, -1);
	TREE T(N, NODE());
	while(--N){
		int u,v,w; cin>>u>>v>>w; u--,v--,w&=1;
		T[u].emplace_back(v,w);
		T[v].emplace_back(u,w);
	}
	dfs(T, dist, 0, 0);
	for(int a:dist) cout<< a << endl;
}
### Prompt

Please provide a cpp coded solution 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
// D - Even Relation
#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using NODE = vector<pair<int,int>>;
using TREE = vector<NODE>;

void dfs(TREE &T, vi &dist, int p, int w){
	if(dist[p] >= 0) return;
	dist[p] = w;
	for(auto c:T[p]){
		int np,nw; tie(np,nw) = c;
		dfs(T, dist, np, w^nw);
	}
}

int main(){
	int N; cin>>N;
	vi dist(N, -1);
	TREE T(N, NODE());
	while(--N){
		int u,v,w; cin>>u>>v>>w; u--,v--,w&=1;
		T[u].emplace_back(v,w);
		T[v].emplace_back(u,w);
	}
	dfs(T, dist, 0, 0);
	for(int a:dist) cout<< a << endl;
}
```