#include <bits/stdc++.h>
#define nmax 100005

using namespace std;

vector <pair <int,int>> g[nmax];
int color[nmax];

void solve(int u, int par, int dist){
    color[u]=dist;
    for(auto v:g[u]){
        if(v.first!=par){
            solve(v.first,u,(dist+v.second)&1);
        }
    }
}

int main(){
    int n;
    cin >> n;
    for(int i=0,u,v,w;i<n-1;i++){
        cin >> u >> v >> w;
        u--,v--;
        g[u].emplace_back(v,w);
        g[v].emplace_back(u,w);
    }
    solve(0,-1,0);
    for(int i=0;i<n;i++){
        cout << color[i] << endl;
    }
    return 0;
}

### 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>
#define nmax 100005

using namespace std;

vector <pair <int,int>> g[nmax];
int color[nmax];

void solve(int u, int par, int dist){
    color[u]=dist;
    for(auto v:g[u]){
        if(v.first!=par){
            solve(v.first,u,(dist+v.second)&1);
        }
    }
}

int main(){
    int n;
    cin >> n;
    for(int i=0,u,v,w;i<n-1;i++){
        cin >> u >> v >> w;
        u--,v--;
        g[u].emplace_back(v,w);
        g[v].emplace_back(u,w);
    }
    solve(0,-1,0);
    for(int i=0;i<n;i++){
        cout << color[i] << endl;
    }
    return 0;
}

```