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:
def examA():
    S = SI()
    if len(S)%2==1:
        print("No")
        return
    for i in range(len(S)//2):
        if S[i*2:i*2+2]!="hi":
            print("No")
            return
    print("Yes")
    return

def examB():
    a, b, M = LI()
    A = LI()
    B = LI()
    D = [LI()for _ in range(M)]
    ans = min(A) + min(B)
    for x,y,c in D:
        cur = A[x-1]+B[y-1]-c
        ans = min(ans,cur)
    print(ans)
    return

def examC():
    def bfs(n, e, fordfs):
        # 点の数、スタートの点、有向グラフ
        W = [-1] * n
        # 各点の状態量、最短距離とか,見たかどうかとか
        W[e] = 0
        que = deque()
        que.append(e)
        while que:
            now = que.popleft()
            nowW = W[now]
            for ne in fordfs[now]:
                if W[ne] == -1:
                    W[ne] = nowW + 1
                    que.append(ne)
        return W
    N = I()
    V = [[]for _ in range(N)]
    for _ in range(N-1):
        a, b = LI()
        a -= 1
        b -= 1
        V[a].append(b)
        V[b].append(a)
    L = bfs(N,0,V)
    #print(L)
    odd = 0
    for l in L:
        if l==-1:
            print(-1)
            return
        if l%2==1:
            odd += 1
    G = []
    if odd*2<N:
        for i in range(N):
            if L[i]%2==1:
                G.append(i)
    else:
        for i in range(N):
            if L[i]%2==0:
                G.append(i)

    used = [False]*(N+1)
    ans = [0]*N
    if N//3>=len(G):
        cur = 3
        for g in G:
            ans[g] = cur
            used[cur] = True
            cur += 3
        cur = 1
        for i in range(N):
            if ans[i]!=0:
                continue
            while(used[cur]):
                cur += 1
            ans[i] = cur
            used[cur] = True
    else:
        cur = 1
        for g in G:
            ans[g] = cur
            used[cur] = True
            cur += 3
            if cur>N:
                cur = 3
        cur = 1
        for i in range(N):
            if ans[i]!=0:
                continue
            while(used[cur]):
                cur += 1
            ans[i] = cur
            used[cur] = True
    print(" ".join(map(str,ans)))
    return

def examD():
    ans = 0
    print(ans)
    return

def examE():
    ans = 0
    print(ans)
    return

def examF():
    ans = 0
    print(ans)
    return

import sys,bisect,itertools,heapq,math,random
from copy import deepcopy
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = 10**(-12)
alphabet = [chr(ord('a') + i) for i in range(26)]

sys.setrecursionlimit(10**6)

if __name__ == '__main__':
    examC()

"""

"""