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;
  map<int, set<pair<int, int>>> mp;
  for (int i=0; i<N-1; i++){
    int u, v, w;
    cin >> u >> v >> w;
    mp[u].insert(make_pair(v,w));
    mp[v].insert(make_pair(u,w));
  }
  vector<long long> dst(N+1,-1);
  queue<int> q;
  q.push(1);
  dst[1] = 0;
  while (!q.empty()){
    int check = q.front();
    q.pop();
    for (auto k: mp[check]){
      if (dst[k.first]<0){
        q.push(k.first);
        dst[k.first] = dst[check]+k.second;
      }
    }
  }
  for (int i=1; i<=N; i++){
    cout << dst[i]%2 << endl;
  }
  return 0;
}