#include <bits/stdc++.h>
using namespace std;
#define el '\n'
bool vis[100005];
int col[100005];
vector<pair<int, int>> adj[100005];

void dfs(int u) {
  vis[u] = true;
  for (auto it: adj[u]) {
    int v = it.first, dis = it.second;
    if (vis[v]) continue;
    col[v] = (col[u] + dis)%2;
    dfs(v);
  }
}
int main() {
  int n;
  cin >> n;
  int sz = n;
  n--;
  while (n--) {
    int u, v, w;
    cin >> u >> v >> w;
    w %= 2;
    adj[u].push_back({v, w});
    adj[v].push_back({u, w});
  }
  col[1] = 0;
  dfs(1);
  for (int i = 1; i <= sz; i++) {
    cout << col[i] << el;
  }
  return 0;
}
### Prompt

Your task is to create 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;
#define el '\n'
bool vis[100005];
int col[100005];
vector<pair<int, int>> adj[100005];

void dfs(int u) {
  vis[u] = true;
  for (auto it: adj[u]) {
    int v = it.first, dis = it.second;
    if (vis[v]) continue;
    col[v] = (col[u] + dis)%2;
    dfs(v);
  }
}
int main() {
  int n;
  cin >> n;
  int sz = n;
  n--;
  while (n--) {
    int u, v, w;
    cin >> u >> v >> w;
    w %= 2;
    adj[u].push_back({v, w});
    adj[v].push_back({u, w});
  }
  col[1] = 0;
  dfs(1);
  for (int i = 1; i <= sz; i++) {
    cout << col[i] << el;
  }
  return 0;
}
```