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(10 ** 6)

n = int(input())

g = [[] for _ in range(n)]

ans = [-1] * n
odd = []
even = []
nums = [[], [], []]
for i in range(1, n + 1):
    nums[i % 3].append(i)

def dfs(v, p = -1, d = 0):
    if d % 2:
        odd.append(v)
    else:
        even.append(v)

    for nv in g[v]:
        if nv == p:
            continue
        dfs(nv, v, d + 1)

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

dfs(0)

zerocnt = len(nums[0])

if len(even) <= zerocnt:
    for even_node in even:
        ans[even_node] = nums[0].pop()
    rest = nums[0] + nums[1] + nums[2]
    for odd_node in odd:
        ans[odd_node] = rest.pop()
elif len(odd) <= zerocnt: 
    for odd_node in odd:
        ans[odd_node] = nums[0].pop()
    rest = nums[0] + nums[1] + nums[2]
    for even_node in even:
        ans[even_node] = rest.pop()
else: 
    for odd_node in odd:
        if len(nums[1]) != 0:
            ans[odd_node] = nums[1].pop()
        else:
            ans[odd_node] = nums[0].pop()
    for even_node in even:
        if len(nums[2]) != 0:
            ans[even_node] = nums[2].pop()
        else:
            ans[even_node] = nums[0].pop()

print(*ans)