#include"bits/stdc++.h"
using namespace std;
#define LL long long
#define MP make_pair
#define PB push_back

int N;
vector<pair<int,int>> edge[100000];

int ans[100000] = {};

void solve(int now, int par, int t){
	ans[now] = t+1;
	for(int i=0; i<edge[now].size(); i++){
		if(edge[now][i].first == par) continue;
		solve(edge[now][i].first, now, (edge[now][i].second%2 == 0 ? t : (t+1)%2));
	}
}

int main(){
	cin >> N;
	for(int i=0; i<N-1; i++){
		int a, b, w;
		cin >> a >> b >> w;
		edge[a-1].PB(MP(b-1,w));
		edge[b-1].PB(MP(a-1,w));
	}
	solve(0,-1,0);
	for(int i=0; i<N; i++) cout << ans[i]-1 << endl;
	
	return 0;
}

### Prompt

In Cpp, your task is to solve 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;
#define LL long long
#define MP make_pair
#define PB push_back

int N;
vector<pair<int,int>> edge[100000];

int ans[100000] = {};

void solve(int now, int par, int t){
	ans[now] = t+1;
	for(int i=0; i<edge[now].size(); i++){
		if(edge[now][i].first == par) continue;
		solve(edge[now][i].first, now, (edge[now][i].second%2 == 0 ? t : (t+1)%2));
	}
}

int main(){
	cin >> N;
	for(int i=0; i<N-1; i++){
		int a, b, w;
		cin >> a >> b >> w;
		edge[a-1].PB(MP(b-1,w));
		edge[b-1].PB(MP(a-1,w));
	}
	solve(0,-1,0);
	for(int i=0; i<N; i++) cout << ans[i]-1 << endl;
	
	return 0;
}

```