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<bits/stdc++.h>

using namespace std;

/*

        Bismillahir Rahmanir Rahim
        Problem :
        Problem Link :
        Topics :
        Solver : Masud Parves
        I Love Code More than Sharks Love Blood <3
*/


#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define SZ(a)           (int)a.size()
#define all(a)  a.begin(), a.end()
#define allr(a)  a.rbegin(), a.rend()


#define sf(a)           scanf("%d",&a)
#define sff(a,b)        scanf("%d %d",&a,&b)
#define sfff(a,b,c)     scanf("%d %d %d",&a,&b,&c)

#define f0(i,b) for(int i=0;i<(b);i++)
#define f1(i,b) for(int i=1;i<=(b);i++)
#define f2(i,a,b) for(int i=(a);i<=(b);i++)
#define fr(i,b,a) for(int i=(b);i>=(a);i--)

#define CIN ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL);
#define TEST_CASE(t) for(int zz=1 ; zz<=t ; zz++)
#define PRINT_CASE printf("Case %d: ",zz)
#define Debug(x)            cout<<#x " = "<<(x)<<endl
#define NOT_VISITED 0
#define IS_VISITED 1
#define WHITE 0
#define GRAY 1
#define BLACK 2



int fx[4]= {1,-1,0,0};
int fy[4]= {0,0,1,-1};

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N  (1<<pos);}
//int Reset(int N,int pos){return N= N & ~(1<<pos);}
//bool Check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/


const double PI = acos(-1.0);
const double EPS = 1e-6;
const int MOD = (int)1e9+7;
const int maxn = (int)2e5+5;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef pair<ll, int> plli;
typedef pair<int, ll> pill;

vector < int > g[maxn];
vector < ll > cost[maxn];
bool visited[maxn];
int color[maxn];

void bfs(int src){
    queue < int >Q;

    Q.push(src);
    visited[src]=1;
    color[src]=1;

    while(!Q.empty()){
        int u=Q.front();
        Q.pop();

        for(int i=0 ; i<g[u].size() ; i++){
            int v=g[u][i];
            if(!visited[i]){
                int w=cost[u][i];
                if(w%2==0){
                    color[v]=color[u];
                }else {
                    color[v]= 1- color[u];
                }
            }
        }
    }
}


int main() {
    CIN
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);

    int n;
    cin >> n;


    f1(i,n-1){
        int u,v;
        ll w;
        cin >> u >> v >> w;

        g[u].pb(v);
        g[v].pb(u);
        cost[u].pb(w);
        cost[v].pb(w);
    }

    bfs(1);
    for(int i=0 ; i<n ; i++){
        cout << color[i] << endl;
    }

    return 0;
}