#include <bits/stdc++.h>
using namespace std;
using PP = pair<int, int>;

int n;
vector<PP> G[100000];
bool col[100000];

void rec(int from, int prev, bool c) {
  col[from] = c;
  for (PP e : G[from]) {
    int to = e.first;
    int dist = e.second;
    if (to == prev) continue;
    rec(to, from, dist % 2 == 0 ? c : !c);
  }
}

int main() {
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    --a; --b;
    G[a].push_back(PP(b, c));
    G[b].push_back(PP(a, c));
  }
  
  rec(0, -1, 0);
  
  for (int i = 0; i < n; ++i) {
    cout << col[i] << endl;
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
using PP = pair<int, int>;

int n;
vector<PP> G[100000];
bool col[100000];

void rec(int from, int prev, bool c) {
  col[from] = c;
  for (PP e : G[from]) {
    int to = e.first;
    int dist = e.second;
    if (to == prev) continue;
    rec(to, from, dist % 2 == 0 ? c : !c);
  }
}

int main() {
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    --a; --b;
    G[a].push_back(PP(b, c));
    G[b].push_back(PP(a, c));
  }
  
  rec(0, -1, 0);
  
  for (int i = 0; i < n; ++i) {
    cout << col[i] << endl;
  }
}

```