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

int N;
vector<pair<int, int>> edges[100000];
int ans[100000];

void dfs(int i, int par){
    for(auto& p : edges[i]) if(p.first != par){
        ans[p.first] = (ans[i] + p.second) % 2;
        dfs(p.first, i);
    }
}

int main(){
    cin >> N;
    for(int i=0; i<N-1; i++){
        int a, b, c;
        cin >> a >> b >> c;
        edges[a-1].push_back({b-1, c%2});
        edges[b-1].push_back({a-1, c%2});
    }

    dfs(0, -1);
    for(int i=0; i<N; i++) cout << ans[i] << endl;
    return 0;
}

### Prompt

Your task is to create 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;
vector<pair<int, int>> edges[100000];
int ans[100000];

void dfs(int i, int par){
    for(auto& p : edges[i]) if(p.first != par){
        ans[p.first] = (ans[i] + p.second) % 2;
        dfs(p.first, i);
    }
}

int main(){
    cin >> N;
    for(int i=0; i<N-1; i++){
        int a, b, c;
        cin >> a >> b >> c;
        edges[a-1].push_back({b-1, c%2});
        edges[b-1].push_back({a-1, c%2});
    }

    dfs(0, -1);
    for(int i=0; i<N; i++) cout << ans[i] << endl;
    return 0;
}

```