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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <fstream>
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <sstream>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <random>
#include <numeric>

using namespace std;

#define INT_MAX_VALUE 2147483647
#define LONG_LONG_MAX_VALUE 9223372036854775807
#define ll long long
#define ld long double

struct XX{
    int a;
    int i;
};

class xxGreater {
public:
    bool operator()(const XX& riLeft, const XX& riRight) const {
        //第2条件
        if((riLeft.a) == (riRight.a)){
            return riLeft.i > riRight.i;//<:昇順(小さいものから順番)、>:降順(大きいものから順番)
            //プライオリティキューの場合は > で、top()すると値の小さいものがとれる
        }
        //第1条件
        return (riLeft.a) > (riRight.a);
    }
};
class xxGreater2 {
public:
    bool operator()(const XX& riLeft, const XX& riRight) const {
        return (riLeft.i) < (riRight.i);
    }
};
class xxGreater3 {
public:
    bool operator()(const XX& riLeft, const XX& riRight) const {
        return (riLeft.a) < (riRight.a);
    }
};


//map<long long,long long> prime_f(long long n){
//    map<long long,long long>res;
//    for(int i=2;i*i<=n;i++){
//        while(n%i==0){
//            ++res[i];
//            n/=i;
//        }
//    }
//    if(n!=1)res[n]=1;
//    return res;
//}

//int n;
////int dat[2*10000000];
////int dat2[2*10000000];
//int dat[10];
//int dat2[10];
//
//void init(int n_){
//    n=1;
//    while(n<n_)n*=2;
//    for(int i=0;i<2*n-1;i++){
//        dat[i]=0;
//        dat2[i]=0;
//    }
//}
//
//void initset(int k,int a){
//    k+=n-1;
//    dat[k]=a;
//    while(k>0){
//        k=(k-1)/2;
//        dat[k]=dat[k*2+1]+dat[k*2+2];
//    }
//}
//
////[a,b)の間を[l,r]区間で比較しアップデート
////引数のindexに注意
////nは固定。initで計算すみ
////update2(L[i],R[i]+1,0,0,n,D[i]);
//void update2(int a,int b,int k,int l,int r,int v){//v更新値、区間は0-index
//    if(r<=a || b<=l)return;
//    if(a<=l && r<=b){
//        dat[k]+=dat2[k];
//        if(r-l>1){
//            dat2[k*2+1]+=dat2[k]/2;
//            dat2[k*2+1]+=dat2[k]/2;
//        }
//        dat2[k]=v*(r-l);
//        return;
//    }else{
//        update2(a,b,k*2+1,l,(l+r)/2,v);
//        update2(a,b,k*2+2,(l+r)/2,r,v);
//        return;
//    }
//}
//
//int query(int a,int b,int k,int l,int r){
//    if(r<=a || b<=l)return 0;
//    if(a<=l && r<=b){
//        dat[k]+=dat2[k];
//        if(r-l>1){
//            dat2[k*2+1]+=dat2[k]/2;
//            dat2[k*2+1]+=dat2[k]/2;
//        }
//        dat2[k]=0;
//        return dat[k];
//    }
//    else{
//        int vl=query(a,b,k*2+1,l,(l+r)/2);
//        int vr=query(a,b,k*2+2,(l+r)/2,r);
//        return vl+vr;
//    }
//}




ll ppar[100001]; //親(1-index)
ll rrank[100001];//木の深さ
ll ssize[100001];

void init(ll n){//
    for(ll i=1;i<=n;i++){
        ppar[i]=i;
        rrank[i]=0;
        ssize[i]=1;
    }
}

ll find(ll x){
    if(ppar[x]==x){
        return x;
    }else{
        return ppar[x]=find(ppar[x]);
    }
}

void unite(ll x,ll y){
    x=find(x);
    y=find(y);
    if(x==y){
        return;
    }
    if(rrank[x]<rrank[y]){
        ppar[x]=y;
        ssize[y]+=ssize[x];//親のサイズの方にサイズを加算
    }else{
        ppar[y]=x;
        ssize[x]+=ssize[y];
        if(rrank[x]==rrank[y]){
            rrank[x]++;
        }
    }
}
bool same(ll x,ll y){
    return find(x)==find(y);
}
ll getsize(ll x){
    return ssize[find(x)];
}

int main(int argc, const char * argv[])
{
    //scanf("%s",S);
    //scanf("%d",&N);
    //scanf("%lld %lld",&target1,&target2);
    //sscanf(tmp.c_str(),"%dd%d%d",&time[i], &dice[i], &z[i]);
    //getline(cin, target);
    //ifstream ifs("a.txt");//テスト用
    //ifs >> a;
    //ここから
    
    //入力高速化
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    int n;
    cin>>n;

    init(n);
    for(int i=0;i<n-1;i++){
        ll u,v,w;
        cin >>u>>v>>w;
        if(w%2==0){
            unite(u,v);
        }
    }
    ll index=1;
    for(ll i=1;i<=n;i++){
        if(getsize(i)>1){
            index=i;
            break;
        }
    }
    for(ll i=1;i<=n;i++){
        if(same(index,i)){
            cout << 0 << endl;
        }else{
            cout << 1 << endl;
        }
    }
    
    
    
    //ここまで
    //cout << "ans" << endl;
    //printf("%.0f\n",ans);//小数点以下表示なし
    //printf("%.7f\n",p);
    
    return 0;
}