#include <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 5;

typedef long long ll;

typedef pair <int, ll> pll;

vector <pll> g[N];
int n;
ll d[N];
int a[N];

void dfs(int u, int p, ll dis) {
  d[u] = dis;
  if(d[u] & 1) {
    a[u] = 1;
  } else {
    a[u] = 0;
  }
  for(auto v : g[u]) {
    if(v.first == p) continue;
    dfs(v.first, u, dis + v.second);
  }
}

int main() {
  cin >> n;
  for(int i = 1; i < n; i++) {
    int u, v; cin >> u >> v;
    ll w; cin >> w;
    g[u].push_back({v, w});
    g[v].push_back({u, w});
  }
  dfs(1, 1, 0);
  for(int i = 1; i <= n; i++) {
    cout << a[i] << '\n';
  }
}
### 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;

const int N = 1e5 + 5;

typedef long long ll;

typedef pair <int, ll> pll;

vector <pll> g[N];
int n;
ll d[N];
int a[N];

void dfs(int u, int p, ll dis) {
  d[u] = dis;
  if(d[u] & 1) {
    a[u] = 1;
  } else {
    a[u] = 0;
  }
  for(auto v : g[u]) {
    if(v.first == p) continue;
    dfs(v.first, u, dis + v.second);
  }
}

int main() {
  cin >> n;
  for(int i = 1; i < n; i++) {
    int u, v; cin >> u >> v;
    ll w; cin >> w;
    g[u].push_back({v, w});
    g[v].push_back({u, w});
  }
  dfs(1, 1, 0);
  for(int i = 1; i <= n; i++) {
    cout << a[i] << '\n';
  }
}
```