Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

3
1 2 2
2 3 1

Sample Output 1

0
0
1

Sample Input 2

5
2 5 2
2 3 10
1 3 8
3 4 2

Sample Output 2

1
0
1
0
1

Solution in C++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef std::pair<int,int> pair2;
typedef std::pair<int,pair2> pair3;
const ll INF = 1LL<<60;


struct UnionFind {
    vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

    UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
        for(int i = 0; i < N; i++) par[i] = i;
    }

    int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }
  
   void unite_l(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx%2 != ry%2) return; //xとyの根が同じ(=違う木にある)時はそのまま
        par[rx] = (ry+1)+100000; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
   }

    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};



int main(){
ll N;
  cin>>N;
  ll u[N],v[N],w[N];
  UnionFind tree(N+1);
 vector<pair<ll,ll>> wro;
  
  for(ll i=1;i<N;i++){
  cin>>u[i]>>v[i]>>w[i];
    //cout<<u[i]<<" "<<v[i]<<" "<<w[i]<<endl;
   if(w[i]%2==0)tree.unite(u[i],v[i]);
   if(w[i]%2!=0)wro.push_back(make_pair(u[i],v[i])); 
  }
  
for(ll i=0;i<(ll)wro.size();i++){
  tree.unite_l(wro[i].first,wro[i].second);
  //cout<<wro[i].first<<","<<wro[i].second<<endl;
}
  
  
for(ll i=1;i<N+1;i++){
cout<<tree.par[i]%2<<endl;
}

 
}