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

int N;
vector<vector<pair<int, int>>> G;
vector<int> V;
vector<int> ans;

void dfs(int v, int dist) {
  V.at(v) = 1;
  ans.at(v) = dist % 2;
  // if (accumulate(V.begin(), V.end(), 0L) == N) return;
  for (auto g : G.at(v)) {
    if (V.at(g.first)) continue;
    dfs(g.first, dist + g.second);
  }
}

int main() {
  cin >> N;
  G.resize(N);
  V.resize(N);
  ans.resize(N);
  for (int a, b, c; cin >> a >> b >> c;) {
    G.at(--a).push_back({--b, c % 2});
    G.at(b).push_back({a, c % 2});
  }
  dfs(0, 0);
  for (auto a : ans) cout << a << "\n";
}
### Prompt

Please formulate a cpp solution 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 <bits/stdc++.h>
using namespace std;

int N;
vector<vector<pair<int, int>>> G;
vector<int> V;
vector<int> ans;

void dfs(int v, int dist) {
  V.at(v) = 1;
  ans.at(v) = dist % 2;
  // if (accumulate(V.begin(), V.end(), 0L) == N) return;
  for (auto g : G.at(v)) {
    if (V.at(g.first)) continue;
    dfs(g.first, dist + g.second);
  }
}

int main() {
  cin >> N;
  G.resize(N);
  V.resize(N);
  ans.resize(N);
  for (int a, b, c; cin >> a >> b >> c;) {
    G.at(--a).push_back({--b, c % 2});
    G.at(b).push_back({a, c % 2});
  }
  dfs(0, 0);
  for (auto a : ans) cout << a << "\n";
}
```