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

typedef long long ll;

ll N;
vector<vector<pair<ll, ll>>> Adj;
vector<ll> R;

void dfs(ll v, ll par, ll r) {
  R[v] = r;
  for (auto uwe : Adj[v]) {
    ll u = uwe.first, we = uwe.second;
    if (u == par) continue;
    dfs(u, v, (r + we) % 2);
  }
}
int main() {
  cin >> N;
  Adj.resize(N);
  for (ll i = 0; i < N - 1; ++i) {
    ll u, v, we;
    cin >> u >> v >> we;
    --u;
    --v;
    Adj[u].push_back(make_pair(v, we));
    Adj[v].push_back(make_pair(u, we));
  }
  R.resize(N);
  dfs(0, -1, 0);
  for (ll r : R) cout << r << endl;
  return 0;
}

### 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;

typedef long long ll;

ll N;
vector<vector<pair<ll, ll>>> Adj;
vector<ll> R;

void dfs(ll v, ll par, ll r) {
  R[v] = r;
  for (auto uwe : Adj[v]) {
    ll u = uwe.first, we = uwe.second;
    if (u == par) continue;
    dfs(u, v, (r + we) % 2);
  }
}
int main() {
  cin >> N;
  Adj.resize(N);
  for (ll i = 0; i < N - 1; ++i) {
    ll u, v, we;
    cin >> u >> v >> we;
    --u;
    --v;
    Adj[u].push_back(make_pair(v, we));
    Adj[v].push_back(make_pair(u, we));
  }
  R.resize(N);
  dfs(0, -1, 0);
  for (ll r : R) cout << r << endl;
  return 0;
}

```