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 struct{int to,w;}dat;
int main(){
  int n;
  cin>>n;
  vector<vector<dat>> adj(n);
  vector<bool> col(n),vis(n,false);
  for(int i=1;i<n;++i){
    int a,b,c;
    cin>>a>>b>>c;
    a--; b--; c=2-c%2;
    adj[a].push_back({b,c});
    adj[b].push_back({a,c});
  }
  queue<int> que;
  col[0]=false;
  que.push(0);
  while(!que.empty()){
    int q = que.front(); que.pop();
    vis[q]=true;
    for(auto d:adj[q])if(!vis[d.to]){
      col[d.to]=(d.w%2)^(col[q]);
      que.push(d.to);
    }
  }
  for(auto i:col)cout<<i<<endl;
}