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<vector<pair<int,int>>> g(N);
  for(int i=0;i<N-1;i++){
    int a,b,c;
    cin>>a>>b>>c;
    a--,b--;
    g.at(a).push_back(make_pair(b,c%2));
    g.at(b).push_back(make_pair(a,c%2));
  }
  vector<int> r(N,-1);
  r.at(0)=0;
  queue<int> q;
  q.push(0);
  while(!(q.empty())){
    int z=q.front();
    q.pop();
    for(pair<int,int> x:g.at(z)){
      if(r.at(x.first)==-1){
        q.push(x.first);
        r.at(x.first)=r.at(z)+x.second;
      }
    }
  }
  for(int i=0;i<N;i++){
    cout<<r.at(i)%2<<endl;
  }
}