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;
int main(){
  int n;
  cin >> n;
  vector<set<pair<int,long> > >hen(n+1);
  int i;
  int u,v;
  long w;
  for(i=0;i<n-1;i++){
    cin >> u >> v >> w;
    hen[u].insert(make_pair(v,w));
    hen[v].insert(make_pair(u,w));
  }
  vector<int>colors(n+1,-1);
  colors[1] = 0;
  queue<int>que;
  que.push(1);
  while(!que.empty()){
    int tmp = que.front();
    que.pop();
    for(auto j:hen[tmp]){
      if(colors[j.first]==-1)que.push(j.first);
      colors[j.first] = (colors[tmp]+j.second)%2;
    }
  }
  for(i=1;i<=n;i++){
    cout << colors[i] << endl;
  }
}