#include<bits/stdc++.h>
using namespace std;
vector<pair<int,bool>> v[100000];
int c[100000];
void init(int n){
  fill(c,c+n,-1);c[0] = 0;
}
void dfs(int f){
  int next;
  for(int i=0;i<v[f].size();i++){
    next = v[f][i].first;
    if(c[next]==-1){
      c[next] = c[f]^v[f][i].second;dfs(v[f][i].first);
    }
  }
}
int main(){
  int n,i,x,y,z;scanf("%d",&n);
  init(n);
  for(i=0;i<n-1;i++){
    scanf("%d%d%d",&x,&y,&z);x--;y--;
    v[x].push_back(make_pair(y,z%2));
    v[y].push_back(make_pair(x,z%2));
  }
  dfs(0);
  for(i=0;i<n;i++) printf("%d\n",c[i]);
  return 0;
}
### Prompt

Generate a CPP solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
vector<pair<int,bool>> v[100000];
int c[100000];
void init(int n){
  fill(c,c+n,-1);c[0] = 0;
}
void dfs(int f){
  int next;
  for(int i=0;i<v[f].size();i++){
    next = v[f][i].first;
    if(c[next]==-1){
      c[next] = c[f]^v[f][i].second;dfs(v[f][i].first);
    }
  }
}
int main(){
  int n,i,x,y,z;scanf("%d",&n);
  init(n);
  for(i=0;i<n-1;i++){
    scanf("%d%d%d",&x,&y,&z);x--;y--;
    v[x].push_back(make_pair(y,z%2));
    v[y].push_back(make_pair(x,z%2));
  }
  dfs(0);
  for(i=0;i<n;i++) printf("%d\n",c[i]);
  return 0;
}
```