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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>
#define LL long long
#define INF 1<<31-1
#define LLINF 1<<63-1
#define MIN(x,y) ((x)<(y)?(x):(y))
#define MAX(x,y) ((x)<(y)?(y):(x))
#define MOD 1000000007
#define NUM 100001
void swap(int* x,int* y){
  int tmp=*x;*x=*y;*y=tmp;
}
int par[NUM];//x番目のノードの親ノード x番目のノードが根なら負(要素数*-1)
//int dep[NUM];//x番目のノードの深さ
int UF(){//初期化
  for(int i=0;i<NUM;i++){par[i]=-1;/*dep[i]=0;*/}
}
int Root(int x){//x番目のノードが属する木の根を探す
  if(par[x]<0)return x;
  else return par[x]=Root(par[x]);
}
int Size(int x){//x番目のノードが属する木の要素数を返す
  return -par[Root(x)];
}
bool Same(int x,int y){//x番目のノードとy番目のノードが同じ木に属しているか返す
  return Root(x)==Root(y);
}
void Unite(int x,int y){//2つの木を結合する
  int tmp=x;x=Root(x);y=Root(y);
  if(x!=y){
    if(Size(x)<Size(y))swap(&x,&y);
    par[x]+=par[y];
    par[y]=x;
  }
}

typedef struct{int u;int v;int w;}P;
/*int cmp(const void* x, const void* y) {
  int n=dep[((P*)x)->u]-dep[((P*)y)->u];
  if(n==0)n=dep[((P*)x)->v]-dep[((P*)y)->v];
  return n;
}*/

int main() {
  UF();
  int N,i;
  scanf("%d",&N);
  P edge[N-1];
  int col[N];
  for(i=0;i<N-1;i++)scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);
  //qsort(edge,N-1,sizeof(P),cmp);
  for(i=0;i<N-1;i++){
    edge[i].u--;edge[i].v--;
    edge[i].w%=2;
    Unite(edge[i].u,edge[i].v);
  }
  for(i=0;i<N;i++)col[i]=0;
  for(i=0;i<N-1;i++){
    int u=edge[i].u,v=edge[i].v;
    if(v==par[u])swap(&u,&v);
    col[v]=col[u]+edge[i].w;
    if(col[v]==2)col[v]=0;
  }

  for(i=0;i<N;i++)printf("%d\n",col[i]);
	return 0;
}