#include<bits/stdc++.h>
using namespace std;
int N;
bool bw[100010], used[100010];
struct edge{
    int to, cost;
};

vector<edge> G[100010];
void dfs(int n){
    used[n] = true;
    for(auto u : G[n]){
        if(used[u.to]) continue;
        if(u.cost % 2 == 0) bw[u.to] = bw[n];
        else bw[u.to] = !bw[n];
        dfs(u.to);
    }
}

int main(){
    cin >> N;
    for(int i = 1; i < N; i++){
        int u, v, w;
        cin >> u >> v >> w;
        G[u].push_back(edge{v,w});
        G[v].push_back(edge{u,w});
    }
    dfs(1);
    for(int i = 1; i <= N; i++){
        cout << bw[i] << endl;
    }
    return 0;
}
### Prompt

Please formulate 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;
int N;
bool bw[100010], used[100010];
struct edge{
    int to, cost;
};

vector<edge> G[100010];
void dfs(int n){
    used[n] = true;
    for(auto u : G[n]){
        if(used[u.to]) continue;
        if(u.cost % 2 == 0) bw[u.to] = bw[n];
        else bw[u.to] = !bw[n];
        dfs(u.to);
    }
}

int main(){
    cin >> N;
    for(int i = 1; i < N; i++){
        int u, v, w;
        cin >> u >> v >> w;
        G[u].push_back(edge{v,w});
        G[v].push_back(edge{u,w});
    }
    dfs(1);
    for(int i = 1; i <= N; i++){
        cout << bw[i] << endl;
    }
    return 0;
}
```