#include <bits/stdc++.h>
using namespace std;

vector<set<pair<int,int>>> X(100000);
vector<int> Bl(100000,5);

void dfs(int x,int y){
  auto it=X[x].begin();
  while(it!=X[x].end()){
    int a=(*it).first;
    int c=(*it).second;
    it++;
    if(a==y) continue;
    Bl[a]=(Bl[x]+c)%2;
    dfs(a,x);
  }
  return;
}

int main(){
  int N;
  cin>>N;
  int a,b,c;
  for(int i=0;i<N-1;i++){
    cin>>a>>b>>c; c%=2;
    a--; b--;
    X[a].insert({b,c});
    X[b].insert({a,c});
  }
  Bl[0]=0;
  dfs(0,512411);
  for(int i=0;i<N;i++){
    cout<<Bl[i]<<endl;
  }
}
### Prompt

Construct a cpp code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

vector<set<pair<int,int>>> X(100000);
vector<int> Bl(100000,5);

void dfs(int x,int y){
  auto it=X[x].begin();
  while(it!=X[x].end()){
    int a=(*it).first;
    int c=(*it).second;
    it++;
    if(a==y) continue;
    Bl[a]=(Bl[x]+c)%2;
    dfs(a,x);
  }
  return;
}

int main(){
  int N;
  cin>>N;
  int a,b,c;
  for(int i=0;i<N-1;i++){
    cin>>a>>b>>c; c%=2;
    a--; b--;
    X[a].insert({b,c});
    X[b].insert({a,c});
  }
  Bl[0]=0;
  dfs(0,512411);
  for(int i=0;i<N;i++){
    cout<<Bl[i]<<endl;
  }
}
```