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:

"""
Writer:SPD_9X2

木を2色で塗り分ける(交互に)

mod3で、
赤/青
10/20
20/10
120/0
0/120s
の4通りの分け方を試せばよい

絶対OKな証明
入る場合→入れればよい
入らない場合→mod0は必ずmod1,mod2以下の数。
mod0を入れても空きがあるとまずいけど、差は1以下なのでその場合はmod1が必ず入る
→おわり

"""


import sys
from collections import deque
def NC_Dij(lis,start):

    ret = [float("inf")] * len(lis)
    ret[start] = 0
    
    q = deque([start])
    plis = [i for i in range(len(lis))]

    while len(q) > 0:
        now = q.popleft()

        for nex in lis[now]:

            if ret[nex] > ret[now] + 1:
                ret[nex] = ret[now] + 1
                plis[nex] = now
                q.append(nex)

    return ret,plis

N = int(input())

lis = [[] for i in range(N)]
for i in range(N-1):

    a,b = map(int,input().split())
    a -= 1
    b -= 1

    lis[a].append(b)
    lis[b].append(a)

dis,plis = NC_Dij(lis,0)

use = [1,2,3]

flag = True
ans = []
for i in range(N):

    nex = use[dis[i] % 2]
    use[dis[i] % 2] += 3
    if nex > N:
        nex = use[2]
        use[2] += 3

    if nex > N:
        flag = False
        break
    ans.append(nex)

if flag:
    print (" ".join(map(str,ans)))
    sys.exit()

use = [2,1,3]
flag = True
ans = []
for i in range(N):

    nex = use[dis[i] % 2]
    use[dis[i] % 2] += 3
    if nex > N:
        nex = use[2]
        use[2] += 3

    if nex > N:
        flag = False
        break
    ans.append(nex)

if flag:
    print (" ".join(map(str,ans)))
    sys.exit()

use = [3,1,2]
flag = True
ans = []
for i in range(N):

    nex = use[dis[i] % 2]
    use[dis[i] % 2] += 3
    if nex > N and nex % 3 == 1:
        nex = use[2]
        use[2] += 3
    if nex > N and nex % 3 == 2:
        nex = use[0]
        use[0] += 3

    if nex > N:
        flag = False
        break
    ans.append(nex)

if flag:
    print (" ".join(map(str,ans)))
    sys.exit()


use = [1,3,2]
flag = True
ans = []
for i in range(N):

    nex = use[dis[i] % 2]
    use[dis[i] % 2] += 3

    if nex > N and nex % 3 == 1:
        nex = use[2]
        use[2] += 3
    if nex > N and nex % 3 == 2:
        nex = use[1]
        use[1] += 3

    if nex > N:
        flag = False
        break
    ans.append(nex)

if flag:
    print (" ".join(map(str,ans)))
    sys.exit()



print (-1)