Problem: 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.

Examples

Input

3
1 2 2
2 3 1


Output

0
0
1


Input

5
2 5 2
2 3 10
1 3 8
3 4 2


Output

1
0
1
0
1

Solution (in PYTHON3):

Solution:
n=int(input())
l=[[] for i in range(n)]
ans=[-1 for i in range(n)]
for i in range(n-1):
  u,v,w=[int(x) for x in input().rstrip().split()]
  l[u-1].append([v-1,w])
  l[v-1].append([u-1,w])

def bfs():
  ans[0]=0
  que=[0]
  while(que):
    now=que.pop(0)
    for i,j in l[now]:
      if ans[i]==-1:
        if j%2==0:
          ans[i]=ans[now]
        else:
          ans[i]=(ans[now]+1)%2
        que.append(i)

bfs()
for i in ans:print(i)