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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;
#define MAXN 100002

vector <pair <int,int> > adj[MAXN];
bool colors[MAXN];

void dfs(int node, int parent = - 1, long long int w = 0){
  colors[node] = w % 2;
  for(pair <int,int> &i : adj[node]){
    if(i.first == parent)continue;
    dfs(i.first,node, w + i.second);
  }
}
int main(){
  int N;
  cin >> N;
  for(int i = 1, a, b, w; i < N; i++){
    	cin >> a >> b >> w;
    	adj[a].push_back({b,w});
    	adj[b].push_back({a,w});
  }
  dfs(1);
  for(int i = 1; i <= N; i++){
    cout << colors[i] << '\n';
  }
}