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:
n=int(input())
dist=[-1]*n
G=[[] for i in range(n)]

for i in range(n-1):
    a,b=map(int,input().split())
    a-=1;b-=1
    G[a].append(b)
    G[b].append(a)

from collections import deque
queue=deque([[0,i] for i in G[0]])
dist[0]=0
while queue:
    now,to=queue.pop()
    # print(now,to)
    if dist[to]!=-1:continue
    dist[to]=1-dist[now]
    for i in G[to]:
        queue.appendleft([to,i])

# print(dist)
q=sum(dist)
LAST=[0]*n
if n/3 <= q and n/3 <= n-q:
    
    A=[i for i in range(1,n+1,3)] + [0]*n
    

    B=[i for i in range(2,n+1,3)] + [0]*n


    now1=0
    now2=0
    for i in range(n):
        if dist[i]:
            LAST[i]=B[now2]
            now2+=1
        else:
            LAST[i]=A[now1]
            now1+=1
    w=3
    for i in range(n):
      if LAST[i]==0:
        LAST[i]=w
        w+=3
    print(*LAST)
            
elif n/3 >q:
    w=3
    for i in range(n):
        if dist[i]:
            LAST[i]=w
            w+=3
    A=[i for i in range(w,n+1,3)]+[i for i in range(1,n+1) if i%3!=0]
    now=0
    for i in range(n):
        if LAST[i]==0:
            LAST[i]=A[now]
            now+=1
    print(*LAST)
else:
    w=3
    for i in range(n):
        if dist[i]==0:
            LAST[i]=w
            w+=3
    A=[i for i in range(w,n+1,3)]+[i for i in range(1,n+1) if i%3!=0]
    now=0
    for i in range(n):
        if LAST[i]==0:
            LAST[i]=A[now]
            now+=1
    print(*LAST)