#include<bits/stdc++.h>

using namespace std;
using ll = long long;

int n;
vector<vector<pair<int,int>>> hen;
vector<int> col;

void dfs(int v, int par) {
    for(auto i : hen[v]) {
        if(i.first == par)continue;
        col[i.first] = (i.second % 2 ? !col[v] : col[v]);
        dfs(i.first, v);
    }
}

int main()
{
    cin >> n;
    hen.resize(n), col.resize(n);
    for(int i = 0; i < n - 1; i++) {
        int a, b, c; cin >> a >> b >> c; a--, b--;
        hen[a].emplace_back(b, c);
        hen[b].emplace_back(a, c);
    }

    dfs(0, -1);
    for(auto i : col)cout << i << endl;
}
### Prompt

Generate a CPP solution to 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;
using ll = long long;

int n;
vector<vector<pair<int,int>>> hen;
vector<int> col;

void dfs(int v, int par) {
    for(auto i : hen[v]) {
        if(i.first == par)continue;
        col[i.first] = (i.second % 2 ? !col[v] : col[v]);
        dfs(i.first, v);
    }
}

int main()
{
    cin >> n;
    hen.resize(n), col.resize(n);
    for(int i = 0; i < n - 1; i++) {
        int a, b, c; cin >> a >> b >> c; a--, b--;
        hen[a].emplace_back(b, c);
        hen[b].emplace_back(a, c);
    }

    dfs(0, -1);
    for(auto i : col)cout << i << endl;
}
```