#include<bits/stdc++.h>
using namespace std;
const int maxn=100000+10;
typedef pair<int,int> ii;

int n;
vector<ii> a[maxn];
int color[maxn];
int d[maxn]={0};
int avl[maxn]={0};

void DFS(int u){
    avl[u]=1;
    for(int i=0;i<a[u].size();i++){
        int v=a[u][i].first;int w=a[u][i].second;
        if (avl[v]==0){
            d[v]=d[u]^w;
            DFS(v);
        }
    }
}

int main(){
    cin>>n;
    for(int i=1;i<=n-1;i++){
        int u,v,w;
        cin>>u>>v>>w;w=w%2;
        a[u].push_back(ii(v,w));
        a[v].push_back(ii(u,w));
    }
    DFS(1);
    for(int i=1;i<=n;i++) cout<<d[i]<<endl;
}

### Prompt

Your challenge is to write 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;
const int maxn=100000+10;
typedef pair<int,int> ii;

int n;
vector<ii> a[maxn];
int color[maxn];
int d[maxn]={0};
int avl[maxn]={0};

void DFS(int u){
    avl[u]=1;
    for(int i=0;i<a[u].size();i++){
        int v=a[u][i].first;int w=a[u][i].second;
        if (avl[v]==0){
            d[v]=d[u]^w;
            DFS(v);
        }
    }
}

int main(){
    cin>>n;
    for(int i=1;i<=n-1;i++){
        int u,v,w;
        cin>>u>>v>>w;w=w%2;
        a[u].push_back(ii(v,w));
        a[v].push_back(ii(u,w));
    }
    DFS(1);
    for(int i=1;i<=n;i++) cout<<d[i]<<endl;
}

```