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>
#define int long long
using namespace std;
int dx[]={0,1,0,-1};
int dy[]={1,0,-1,0};
signed main(){
  int N;
  cin>>N;
  vector<vector<pair<int,int>>> A(N);
  for(int i=1;i<N;i++){
    int a,b,c;
    cin>>a>>b>>c;
    A[a-1].push_back({b-1,c});
    A[b-1].push_back({a-1,c});
  }
  vector<int> B(N,-1);
  queue<int> Q;
  B[0]=0;
  Q.push(0);
  while(!Q.empty()){
    int I=Q.front();
    Q.pop();
    for(pair<int,int> p:A[I])
      if(B[p.first]==-1){
        B[p.first]=(B[I]+p.second)%2;
        Q.push(p.first);
      }
  }
  for(int i=0;i<N;i++)cout<<B[i]<<endl;
}