#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cmath>
#define REP(i, n) for (int i = 0; i < (n); i++)

using namespace std;

vector<pair<int, int>> G[100000];
int dist[100000];

void dfs(int u, int p) {
  for (auto e : G[u]) if (e.first != p) {
    dist[e.first] = (dist[u] + e.second) % 2;
    dfs(e.first, u);
  }
}

int main() {
  int N;
  cin >> N;
  REP(i, N - 1) {
    int u, v, w;
    cin >> u >> v >> w;
    u--; v--;
    G[u].emplace_back(v, w);
    G[v].emplace_back(u, w);
  }
  dfs(0, -1);
  for (int i = 0; i < N; i++) {
    cout << dist[i] % 2 << '\n';
  }
}

### 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 <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cmath>
#define REP(i, n) for (int i = 0; i < (n); i++)

using namespace std;

vector<pair<int, int>> G[100000];
int dist[100000];

void dfs(int u, int p) {
  for (auto e : G[u]) if (e.first != p) {
    dist[e.first] = (dist[u] + e.second) % 2;
    dfs(e.first, u);
  }
}

int main() {
  int N;
  cin >> N;
  REP(i, N - 1) {
    int u, v, w;
    cin >> u >> v >> w;
    u--; v--;
    G[u].emplace_back(v, w);
    G[v].emplace_back(u, w);
  }
  dfs(0, -1);
  for (int i = 0; i < N; i++) {
    cout << dist[i] % 2 << '\n';
  }
}

```