#include <vector>
#include <iostream>
using namespace std;
int col[1001000];
vector<pair<int, int> > adj[100100];

void dfs(int i, int p, int c) {
    col[i] = c;
    for (auto e : adj[i]) {
        int j = e.first, w = e.second;
        if (j != p) dfs(j, i, c^w);
    }
}

int main() {
    int n; cin >> n;
    for (int i = 1; i < n; i++) {
        int u, v, w; cin >> u >> v >> w;
        u--; v--; w %= 2;
        adj[u].push_back({v, w});
        adj[v].push_back({u, w});
    }
    dfs(0, -1, 0);
    for (int i = 0; i < n; i++) cout << col[i] << endl;
}

### 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 <vector>
#include <iostream>
using namespace std;
int col[1001000];
vector<pair<int, int> > adj[100100];

void dfs(int i, int p, int c) {
    col[i] = c;
    for (auto e : adj[i]) {
        int j = e.first, w = e.second;
        if (j != p) dfs(j, i, c^w);
    }
}

int main() {
    int n; cin >> n;
    for (int i = 1; i < n; i++) {
        int u, v, w; cin >> u >> v >> w;
        u--; v--; w %= 2;
        adj[u].push_back({v, w});
        adj[v].push_back({u, w});
    }
    dfs(0, -1, 0);
    for (int i = 0; i < n; i++) cout << col[i] << endl;
}

```