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

vector<vector<pair<int, int>>> to(100000);
vector<int> ans(100000);

void dfs(int v = 0, int p = -1, int c = 0)
{
  ans.at(v) = c;
  for (auto n : to.at(v)) {
    if (n.first == p) continue;
    dfs(n.first, v, n.second xor c);
  }
}

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

  for (int i = 0; i < n - 1; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    u--; v--; w %= 2;
    to.at(u).emplace_back(v, w);
    to.at(v).emplace_back(u, w);
  }

  dfs();

  for (int i = 0; i < n; i++) {
    cout << ans.at(i) << 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;

vector<vector<pair<int, int>>> to(100000);
vector<int> ans(100000);

void dfs(int v = 0, int p = -1, int c = 0)
{
  ans.at(v) = c;
  for (auto n : to.at(v)) {
    if (n.first == p) continue;
    dfs(n.first, v, n.second xor c);
  }
}

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

  for (int i = 0; i < n - 1; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    u--; v--; w %= 2;
    to.at(u).emplace_back(v, w);
    to.at(v).emplace_back(u, w);
  }

  dfs();

  for (int i = 0; i < n; i++) {
    cout << ans.at(i) << endl;
  }
}

```