Problem: We have a tree with N vertices. The vertices are numbered 1 to N, and the i-th edge connects Vertex a_i and Vertex b_i.

Takahashi loves the number 3. He is seeking a permutation p_1, p_2, \ldots , p_N of integers from 1 to N satisfying the following condition:

* For every pair of vertices (i, j), if the distance between Vertex i and Vertex j is 3, the sum or product of p_i and p_j is a multiple of 3.



Here the distance between Vertex i and Vertex j is the number of edges contained in the shortest path from Vertex i to Vertex j.

Help Takahashi by finding a permutation that satisfies the condition.

Constraints

* 2\leq N\leq 2\times 10^5
* 1\leq a_i, b_i \leq N
* The given graph is a tree.

Input

Input is given from Standard Input in the following format:


N
a_1 b_1
a_2 b_2
\vdots
a_{N-1} b_{N-1}


Output

If no permutation satisfies the condition, print `-1`.

Otherwise, print a permutation satisfying the condition, with space in between. If there are multiple solutions, you can print any of them.

Example

Input

5
1 2
1 3
3 4
3 5


Output

1 2 5 4 3

Solution (in PYTHON3):

Solution:
# import sys
# sys.setrecursionlimit(2147483647)

n = int(input())
edges = [[] for _ in range(n)]
for _ in range(n - 1):
  a, b = map(int, input().split())
  edges[a - 1].append(b - 1)
  edges[b - 1].append(a - 1)


vertice = [-1 for _ in range(n)]
stack = [(0, 0)]

while stack:
    parent, vertex = stack.pop()
    vertice[vertex] = vertice[parent] * (-1)
    for next in edges[vertex]:
        if next != parent:
            stack.append((vertex, next))

# vertice = [0 for _ in range(n)]
#
#
# def dfs(v, color_i):
#     vertice[v] = color_i
#     for i in edges[v]:
#         if vertice[i] == 0:
#             dfs(i, color_i*(-1))
#
# dfs(0, 1)

ans = []
if vertice.count(1) <= n // 3:
    not3 = [i for i in range(1, n+1) if i % 3 != 0]
    num3 = [i for i in range(1, n+1) if i % 3 == 0]
    for v in vertice:
        if v == 1:
            if num3:
                ans.append(str(num3.pop()))
            else:
                ans.append(str(not3.pop()))
        else:
            if not3:
                ans.append(str(not3.pop()))
            else:
                ans.append(str(num3.pop()))
elif vertice.count(-1) <= n // 3:
    not3 = [i for i in range(1, n+1) if i % 3 != 0]
    num3 = [i for i in range(1, n+1) if i % 3 == 0]
    for v in vertice:
        if v == -1:
            if num3:
                ans.append(str(num3.pop()))
            else:
                ans.append(str(not3.pop()))
        else:
            if not3:
                ans.append(str(not3.pop()))
            else:
                ans.append(str(num3.pop()))
else:
    num1 = [i for i in range(1, n+1) if i % 3 == 1]
    num2 = [i for i in range(1, n+1) if i % 3 == 2]
    num3 = [i for i in range(1, n+1) if i % 3 == 0]
    for v in vertice:
        if v == 1:
            if num1:
                ans.append(str(num1.pop()))
            else:
                ans.append(str(num3.pop()))
        else:
            if num2:
                ans.append(str(num2.pop()))
            else:
                ans.append(str(num3.pop()))

print(" ".join(ans))