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: Accepted)

#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <list>
#include <time.h>
#include <math.h>
#include <random>
#include <deque>
#include <queue>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <iomanip>
#include <bitset>
#include <sstream>
#include <chrono>
#include <cstring>
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double db;

#define ft first
#define sd second
#define flag bool fl=true

#define mod 1000000007
#define N 100001
#define MX 1000000001
#define MXX 1000000000000000001

#define pb push_back
#define For(i, a, b) for(int i=a;i<=b;i++)
#define T int t, qq=0; cin>>t; while(qq++<t)
#define fast ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0)
#define scarr(array, size) for(int i=0;i<size;i++) cin>>array[i]

#define deg(rad) (rad*180.0)/pi
#define pi acos(-1)
#define log(n) (int)log2(n)
#define logg(a)(b) log(b)/log(a)


/*
//----------useful_functions----------

long long extended_euclid(ll a, ll b, ll &x, ll &y){//ax+by=c
    if(b==0)
    {
        x=1, y=0;
        return a;
    }

    ll x1, y1;
    ll d=extended_euclid(b, a%b, x1, y1);
    x=y1;
    y=x1-y1*(a/b);
    return d;
}
long long bigMod(ll a, ll b){//(a^b)%mod
    ll ans=1;
    while(b)
    {
        if(b%2==1) ans=(ans*a)%mod;
        a=(a*a)%mod;
        b/=2;
    }
    return ans%mod;
}
long long modInverse(ll a, ll b){//a*x=1 mod b
    ll x, y;
    ll gcd=extended_euclid(a, b, x, y);

    if(gcd==1) return ((x%b)+b)%b;
    else return 1;
}
long long nCr(ll n, ll r){//n<=1e6
    //declare globally-> ll fact[1000001];
    fact[1]=1;
    for(ll i=2;i<1000001;i++)
        fact[i]=(i*fact[i-1])%mod;

    return (fact[n]*modInverse(fact[r], mod)*modInverse(fact[n-r], mod))%mod;
}
long long josephus(ll n, ll k){//k'th person is killed out of n in the circle
    if(n==1) return 1;
    else return (josephus(n-1, k)+k-1)%n+1;
}

vector<int> sieve(ll n){
    bool fl[n+1];
    vector<int> primes;
    for(int i=0;i<=n;i++) fl[i]=false;

    for(int i=3;i*i<=n;i+=2)
        for(int j=i+i;j<=n;j+=i)
            fl[j]=true;

    primes.pb(2);
    for(int i=3;i<=n;i+=2) if(!fl[i]) primes.pb(i);
    return primes;
}
*/

int main()
{
    int n;
    cin>>n;
    vector<pair<int, int> > v[n+1];
    vector<ll> vis(n+1, MXX);
    for(int i=0;i<n-1;i++)
    {
        int x, y, z;
        cin>>x>>y>>z;
        v[x].push_back({y, z});
        v[y].push_back({x, z});
    }

    queue<int> q;
    q.push(1);
    vis[1]=0;
    while(!q.empty())
    {
        int x=q.front();
        q.pop();

        for(auto it=v[x].begin();it!=v[x].end();it++)
        {
            if(vis[x]+it->second<vis[it->first])
            {
                vis[it->first]=vis[x]+it->second;
                q.push(it->first);
            }
        }
    }

    for(int i=1;i<=n;i++)
    {
        if(vis[i]%2==0) cout<<0<<endl;
        else cout<<1<<endl;
    }
//        cout<<"Case "<<qq<<": ";
}