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

typedef long long int ll;
#define rep(i,n) for(int i = 0; i < n; i++)

vector<pair<ll,ll>> E[100005];
ll ans[100005];

void dfs(ll fro, ll to, ll dis) {
  ans[to] = dis%2;
  for (auto& e:E[to]) {
    if (e.first != fro) dfs(to, e.first, dis+e.second);
  }
}

int main() {
  int n;
  cin >> n;
  ll u,v,w;
  rep(i,n-1) {
    cin >> u >> v >> w;
    E[u-1].push_back({v-1, w%2});
    E[v-1].push_back({u-1, w%2});
  }
  dfs(-1,0,0);
  rep(i,n) cout << ans[i] << endl;
  return 0;
}

### Prompt

Develop a solution in cpp 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 <bits/stdc++.h>
using namespace std;

typedef long long int ll;
#define rep(i,n) for(int i = 0; i < n; i++)

vector<pair<ll,ll>> E[100005];
ll ans[100005];

void dfs(ll fro, ll to, ll dis) {
  ans[to] = dis%2;
  for (auto& e:E[to]) {
    if (e.first != fro) dfs(to, e.first, dis+e.second);
  }
}

int main() {
  int n;
  cin >> n;
  ll u,v,w;
  rep(i,n-1) {
    cin >> u >> v >> w;
    E[u-1].push_back({v-1, w%2});
    E[v-1].push_back({u-1, w%2});
  }
  dfs(-1,0,0);
  rep(i,n) cout << ans[i] << endl;
  return 0;
}

```