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 Python (3.4.3) (judge verdict: Accepted)

# coding: utf-8
class PotUnionFind:
    def __init__(self, size):
        self.parent = [-1 for i in range(size)]#非負なら親ノード，負ならグループの要素数
        self.diff_p = [0 for i in range(size)]#親ノードを基準としたポテンシャル

    def root(self, x): #root(x): xの根ノードを返す
        if self.parent[x] < 0:
            return x
        else:
            rx = self.root(self.parent[x]) #xをxの根rxに直接つなぎたい
            self.diff_p[x] += self.diff_p[self.parent[x]] #そのために，diff(rx,x)を計算
            self.parent[x] = rx #xをrxに直接つなげる
            return rx

    def weight(self, x):
        self.root(x)
        return self.diff_p[x]

    def merge(self, x, y, dxy): #ポテンシャル差の条件p(y)-p(x)=dxyでxとyのグループをまとめる
        dxy += self.weight(x) - self.weight(y) #dxyをdiff(rx,ry)で置き換え
        x = self.root(x) #rxを新たにxと名づける
        y = self.root(y) #ryを新たにyと名づける
        if x == y:
            return False
        if self.parent[x] > self.parent[y]: #xの要素数がyの要素数より「小さい」とき入れ替える
            x,y,dxy=y,x,-dxy
        self.parent[x] += self.parent[y] #xの要素数を更新
        self.parent[y] = x #yをxにつなぐ
        self.diff_p[y] = dxy #yの相対ポテンシャルを更新
        return True

    def issame(self, x, y): #issame(x,y): xとyが同じグループにあるならTrue
        return self.root(x) == self.root(y)
        
    def diff(self,x,y): #diff(x,y): xを基準としたyのポテンシャルを返す 
        if self.root(x) == self.root(y): #この時点でxの親はroot(x)
            return self.diff_p[y] - self.diff_p[x]
        else:
            return None

    def size(self,x): #size(x): xのいるグループの要素数を返す
        return -self.parent[self.root(x)]

n = int(input())

T=PotUnionFind(n)
for _ in [0]*(n-1):
    u,v,w = [int(i) for i in input().split()]
    T.merge(u-1,v-1,w)

for i in range(n):
    print(T.diff(0,i)%2)