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:
from collections import deque
import random
isdebug=0
if isdebug==0:
    N=int(input())
    G=[[] for i in range(N)]
    for i in range(N-1):
        a,b=map(int,input().split())
        G[a-1].append(b-1)
        G[b-1].append(a-1)
else:
    N=20*10**4
    G=[[] for i in range(N)]
    for i in range(N-1):
        a,b=random.randrange(i+1),i+1
        G[a].append(b)
        G[b].append(a)
dist=[-1 for i in range(N)]
X=[]
Y=[]
q=deque([0])
dist[0]=0
while(len(q)>0):
    r=q.popleft()
    if dist[r]%2==0:
        X.append(r)
    else:
        Y.append(r)
    for p in G[r]:
        if dist[p]==-1:
            dist[p]=dist[r]+1
            q.append(p)
P=[0 for i in range(N)]
que=deque([])
if min(len(X),len(Y))<=N//3:
    for i in range(1,N+1):
        if i%3==0:
            que.appendleft(i)
        else:
            que.append(i)
    if len(X)<len(Y):
        for i in X:
            t=que.popleft()
            P[i]=t
        for i in Y:
            t=que.pop()
            P[i]=t
    else:
        for i in Y:
            t=que.popleft()
            P[i]=t
        for i in X:
            t=que.pop()
            P[i]=t
else:
    for i in range(1,N+1):
        if i%3==0:
            que.appendleft(i)
    for i in range(1,N+1):
        if i%3==1:
            que.appendleft(i)
        elif i%3==2:
            que.append(i)
    for i in X:
        t=que.popleft()
        P[i]=t
    for i in Y:
        t=que.pop()
        P[i]=t
print(*P)