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

vector<int> d;
vector<vector<pair<int, int>>> path;

void dfs(int x, int s){
  s %= 2;
  d[x] = s;

  for(auto next : path[x]){
    if(d[next.first] == -1) dfs(next.first, s + next.second);
  }
}


int main(){
  int n; cin >> n;

  d.resize(n);
  path.resize(n);

  for(int i=0; i<n-1; ++i){
    d[i] = -1;
    int u, v, w; cin >> u >> v >> w;
    --u; --v;
    path[u].push_back(make_pair(v, w));
    path[v].push_back(make_pair(u, w));
  }
  d[n-1] = -1;

  dfs(0, 0);

  for(int i=0; i<n; ++i){
    cout << d[i] << endl;
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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<int> d;
vector<vector<pair<int, int>>> path;

void dfs(int x, int s){
  s %= 2;
  d[x] = s;

  for(auto next : path[x]){
    if(d[next.first] == -1) dfs(next.first, s + next.second);
  }
}


int main(){
  int n; cin >> n;

  d.resize(n);
  path.resize(n);

  for(int i=0; i<n-1; ++i){
    d[i] = -1;
    int u, v, w; cin >> u >> v >> w;
    --u; --v;
    path[u].push_back(make_pair(v, w));
    path[v].push_back(make_pair(u, w));
  }
  d[n-1] = -1;

  dfs(0, 0);

  for(int i=0; i<n; ++i){
    cout << d[i] << endl;
  }
}

```