#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;

vector<ll> dis(1e5,-1);
vector<map<ll,ll>> box(1e5);

void dfs(ll v, ll d) {
  dis[v]=d%2;
  for(auto p: box[v]) {
    if(dis[p.first]<0) dfs(p.first, p.second+d);
  }
}

int main() {
  ll zero=0;
  ll n;
  cin >> n;
  for(int i=0;i<n-1;i++) {
    ll u,v,w;
    cin >> u >> v >> w;
    u--;
    v--;
    box[u][v]=w%2;
    box[v][u]=w%2;
  }
  dfs(0,1);
  for(ll i=0;i<n;i++) {
    cout << dis[i] << endl;
  }
}
### Prompt

Create a solution in Cpp for 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;
typedef pair<ll,ll> pll;

vector<ll> dis(1e5,-1);
vector<map<ll,ll>> box(1e5);

void dfs(ll v, ll d) {
  dis[v]=d%2;
  for(auto p: box[v]) {
    if(dis[p.first]<0) dfs(p.first, p.second+d);
  }
}

int main() {
  ll zero=0;
  ll n;
  cin >> n;
  for(int i=0;i<n-1;i++) {
    ll u,v,w;
    cin >> u >> v >> w;
    u--;
    v--;
    box[u][v]=w%2;
    box[v][u]=w%2;
  }
  dfs(0,1);
  for(ll i=0;i<n;i++) {
    cout << dis[i] << endl;
  }
}
```