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 PYTHON):

Solution:
from Queue import Queue
n = int(raw_input())
adjList = [[] for _ in range(n+1)]
colors = [-1] * (n+1)
start = None
for i in range(n-1):
    u, v, w = map(int, raw_input().split())
    adjList[u].append((v, w))
    adjList[v].append((u, w))
    if start is None:
        start = u
def bfs(adjList, colors, start):
    colors[start] = 0
    q = Queue()
    q.put(start)
    while not q.empty():
        u = q.get()
        for adj, w in adjList[u]:
            if colors[adj] == -1:
                if w % 2 == 0:
                    colors[adj] = colors[u]
                else:
                    colors[adj] = colors[u]^1
                q.put(adj)
bfs(adjList, colors, start)
for i in range(1, n+1):
    if colors[i] != -1:
        print colors[i]