#include<bits/stdc++.h>
using namespace std;

#define LL long long
const int N = 1e5 + 5;
vector<pair<int,int>> g[N];
LL d[N]; int x, y, w;

void dfs(int v, int p) {
    for(auto u : g[v]) {
        if(u.first != p) {
            d[u.first] = d[v] + u.second;
            dfs(u.first, v);
        }
    }
}

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    
    int n; cin >> n;
    for(int i = 0; i < n-1; i++) {
        cin >> x >> y >> w;
        g[x].emplace_back(y, w);
        g[y].emplace_back(x, w);
    }    

    dfs(1, 0);

    for(int i = 1; i <= n; i++) cout << d[i] % 2 << '\n';
}
### 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;

#define LL long long
const int N = 1e5 + 5;
vector<pair<int,int>> g[N];
LL d[N]; int x, y, w;

void dfs(int v, int p) {
    for(auto u : g[v]) {
        if(u.first != p) {
            d[u.first] = d[v] + u.second;
            dfs(u.first, v);
        }
    }
}

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    
    int n; cin >> n;
    for(int i = 0; i < n-1; i++) {
        cin >> x >> y >> w;
        g[x].emplace_back(y, w);
        g[y].emplace_back(x, w);
    }    

    dfs(1, 0);

    for(int i = 1; i <= n; i++) cout << d[i] % 2 << '\n';
}
```