#include <iostream>
#include <vector>
using namespace std;

int n;
vector<pair<int, int>> adj[1000000];
int dst[1000000];

void dfs(int u, int prv, int w){
  dst[u] = w & 1;
  for (int i = 0, nx, l; i < adj[u].size(); i++){
    nx = adj[u][i].first, l = adj[u][i].second;
    if (nx != prv){
      dfs(nx, u, w ^ l);
    }
  }
}
int main(){
  cin >> n;
  for (int i = 1, u, v, w; i < n; i++){
    cin >> u >> v >> w;
    w &= 1;
    adj[u].push_back(make_pair(v, w));
    adj[v].push_back(make_pair(u, w));
  }
  dfs(1, -1, 0);
  for (int i = 1; i <= n; i++) cout << dst[i] << '\n';
  return 0;
}

### Prompt

Please create a solution in Cpp 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 <iostream>
#include <vector>
using namespace std;

int n;
vector<pair<int, int>> adj[1000000];
int dst[1000000];

void dfs(int u, int prv, int w){
  dst[u] = w & 1;
  for (int i = 0, nx, l; i < adj[u].size(); i++){
    nx = adj[u][i].first, l = adj[u][i].second;
    if (nx != prv){
      dfs(nx, u, w ^ l);
    }
  }
}
int main(){
  cin >> n;
  for (int i = 1, u, v, w; i < n; i++){
    cin >> u >> v >> w;
    w &= 1;
    adj[u].push_back(make_pair(v, w));
    adj[v].push_back(make_pair(u, w));
  }
  dfs(1, -1, 0);
  for (int i = 1; i <= n; i++) cout << dst[i] << '\n';
  return 0;
}

```