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;
typedef long long ll;
int main(){
 int n;
 cin >> n;
 vector<pair<int,int>> adj[n+1];
 for(int i=0;i<n-1;i++){
  int u,v,w;
  cin>>u>>v>>w;
  adj[u].push_back({v,w});
  adj[v].push_back({u,w});
 }
 queue<int> q;
 q.push(1);
 vector<int> d(n+1,-1);
 d[1]=0;
 vector<int> col(n+1,0);
 while(!q.empty()){
  int u=q.front();
  if(d[u]&1) col[u]=1;
  q.pop();
  for(auto v:adj[u]){
   if(d[v.first]==-1){
    d[v.first]=d[u]+v.second;
    q.push(v.first);
   }
  }
 }
 for(int i=1;i<=n;i++){
  cout<<col[i]<<endl;
 }
 return 0;
}