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
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")


# サイズが大きいとdefaultdictが早い(for文を書かなくて済む)
from collections import defaultdict
n = int(input())
ns = defaultdict(set)
for _ in range(n-1):
    u,v = map(int, input().split())
    u -= 1
    v -= 1
    ns[u].add(v)
    ns[v].add(u)

def is_bip(ns):
    start = 0
    q = [start]
    cs = [None]*n
    cs[start] = 1
    while q:
        u = q.pop()
        c = cs[u]
        cc = int(not c)
        for v in ns[u]:
            if cs[v] is None:
                cs[v] = cc
                q.append(v)
            elif cs[v]==c:
                return False, None
    return True, cs
_, cs = is_bip(ns)
v1, v0 = 0, 0
for num in cs:
    if num==0:
        v0 += 1
    else:
        v1 += 1
num0 = [i for i in range(1,n+1) if i%3==0]
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]
if v0<=len(num0) or v1<=len(num0):
    if v0<=len(num0):
        val = 0
    else:
        val = 1
    for i in range(n):
        if cs[i]==val:
            cs[i] = num0.pop()
        else:
            if num1:
                cs[i] = num1.pop()
            elif num2:
                cs[i] = num2.pop()
            else:
                cs[i] = num0.pop()
else:
    if v0<=v1:
        val = 0
    else:
        val = 1
    for i in range(n):
        if cs[i]==val:
            if num1:
                cs[i] = num1.pop()
            else:
                cs[i] = num0.pop()
        else:
            if num2:
                cs[i] = num2.pop()
            else:
                cs[i] = num0.pop()
write(" ".join(map(str, cs)))