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

bool seen[110000];
int vec[110000];
map<pair<int,int>,int> ma;

void dfs(const vector<vector<int>> &G,int v){
  seen[v]=true;
  for(auto next_v:G[v]){
    if(seen[next_v])continue;
    vec[next_v]=vec[v]+ma[{v,next_v}];
    dfs(G,next_v);
  }
}

int main(){
  int n;
  cin>>n;
  vector<vector<int>> g(n);
  for(int i=0;i<n-1;i++){
    int a,b,c;
    cin>>a>>b>>c;
    a--;
    b--;
    c%=2;
    g[a].push_back(b);
    g[b].push_back(a);
    ma[{a,b}]=c;
    ma[{b,a}]=c;
  }
  vec[0]=0;
  dfs(g,0);
  for(int i=0;i<n;i++){
    cout<<vec[i]%2<<endl;
  }
}
### Prompt

In cpp, your task is to solve 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;

bool seen[110000];
int vec[110000];
map<pair<int,int>,int> ma;

void dfs(const vector<vector<int>> &G,int v){
  seen[v]=true;
  for(auto next_v:G[v]){
    if(seen[next_v])continue;
    vec[next_v]=vec[v]+ma[{v,next_v}];
    dfs(G,next_v);
  }
}

int main(){
  int n;
  cin>>n;
  vector<vector<int>> g(n);
  for(int i=0;i<n-1;i++){
    int a,b,c;
    cin>>a>>b>>c;
    a--;
    b--;
    c%=2;
    g[a].push_back(b);
    g[b].push_back(a);
    ma[{a,b}]=c;
    ma[{b,a}]=c;
  }
  vec[0]=0;
  dfs(g,0);
  for(int i=0;i<n;i++){
    cout<<vec[i]%2<<endl;
  }
}
```