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=sys.stdin.readline
N=int(input())
P=[0 for i in range(N+1)]

L=[[]for i in range(N+1)]
for i in range(N-1):
    a,b=map(int,input().split())
    L[a].append(b)
    L[b].append(a)
#print(L)

d1=[10**9 for i in range(N+1)]
d1[1]=0
Q=[]
for i in L[1]:
    Q.append([i,1])
#print(Q)
#print(d1)
for i in range(N-1):
    d1[Q[i][0]]=Q[i][1]
    for j in L[Q[i][0]]:
        if d1[j]==10**9:
            Q.append([j,Q[i][1]+1])
#print(d1)
ODD=0
EVE=0
for i in range(1,N+1):
    if d1[i]%2==0:
        EVE+=1
    else:
        ODD+=1
#print(ODD,EVE)
M1=[]
M2=[]
M3=[]
for i in range(1,N+1):
    if i%3==1:
        M1.append(i)
    elif i%3==2:
        M2.append(i)
    else:
        M3.append(i)
if min(ODD,EVE)*2<max(ODD,EVE):
    if ODD<EVE:
        D={}
        ans=[0 for i in range(N+1)]
        cnt=0
        for i in range(1,N+1):
            if d1[i]%2==1:
                ans[i]=M3[cnt]
                D[M3[cnt]]=1
                cnt+=1
        num=1
        for i in range(1,N+1):
            #print(D)
            if d1[i]%2==0:
                if num in D:
                    #print(True)
                    num+=1
                ans[i]=num
                num+=1
    else:
        D={}
        ans=[0 for i in range(N+1)]
        cnt=0
        for i in range(1,N+1):
            if d1[i]%2==0:
                ans[i]=M3[cnt]
                D[M3[cnt]]=1
                cnt+=1
        num=1
        for i in range(1,N+1):
            #print(D)
            if d1[i]%2==1:
                if num in D:
                    #print(True)
                    num+=1
                ans[i]=num
                num+=1
    print(*ans[1:])
else:
    ans=[0 for i in range(N+1)]
    m1=0
    m2=0
    m3=0
    for i in range(1,N+1):
        if d1[i]%2==0:
            if m1<len(M1):
                ans[i]=M1[m1]
                m1+=1
            else:
                ans[i]=M3[m3]
                m3+=1
        else:
            if m2<len(M2):
                ans[i]=M2[m2]
                m2+=1
            else:
                ans[i]=M3[m3]
                m3+=1
    print(*ans[1:])