#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> pii;
vector<pii> adj[100001];
bool color[100001];
void dfs (int cur, int par) {
  for (pii p : adj[cur]) if (p.first != par) {
    if (p.second % 2 == 0) color[p.first] = color[cur];
    else color[p.first] = !color[cur];
    dfs(p.first,cur);
  }
}
int main() {
  int n;scanf("%d",&n);
  for (int x = 0; x < n-1; x++) {
    int a,b,c;
    scanf ("%d %d %d",&a,&b,&c);
    adj[a].push_back({b,c});
    adj[b].push_back({a,c});
  }
  dfs(1,1);
  for (int x = 1; x <= n; x++) printf ("%d\n",color[x]);
  return 0;
}
### 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 pair<int,int> pii;
vector<pii> adj[100001];
bool color[100001];
void dfs (int cur, int par) {
  for (pii p : adj[cur]) if (p.first != par) {
    if (p.second % 2 == 0) color[p.first] = color[cur];
    else color[p.first] = !color[cur];
    dfs(p.first,cur);
  }
}
int main() {
  int n;scanf("%d",&n);
  for (int x = 0; x < n-1; x++) {
    int a,b,c;
    scanf ("%d %d %d",&a,&b,&c);
    adj[a].push_back({b,c});
    adj[b].push_back({a,c});
  }
  dfs(1,1);
  for (int x = 1; x <= n; x++) printf ("%d\n",color[x]);
  return 0;
}
```