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())
a= [list(map(int, input().split())) for i in range(n-1)]

b = [[] for i in range(n)]
a1=[]
a2=[]
ans=[0]*n
for i in range(n-1):
    b[a[i][0]-1].append(a[i][1]-1)
    b[a[i][1]-1].append(a[i][0]-1)

import collections
from collections import deque
def tree(s):

    INF=-10**9
    dis = [INF for i in range(n)]
    dis[s]=0
    def bfs():
        d = deque()
        d.append(s)
        a1.append(s)

        while len(d):
            x = d.popleft()

            for i in range(len(b[x])):
                y=b[x][i]
                if dis[y] == INF:
                    d.append(y)
                    dis[y]=dis[x]+1
                    if dis[y]%2==0:
                        a1.append(y)
                    else:
                        a2.append(y)
        return a1,a2
    return bfs()

tree(0)

j = 0
k = 0
if min(len(a1),len(a2))<=n//3:
    if len(a1)<=n//3:
        for i in range(1,n+1):
            if i<=len(a1)*3 and i%3==0 and j<=len(a1)-1:
                ans[a1[j]]=i
                j+=1
            else:
                ans[a2[k]]=i
                k+=1
    else:
        for i in range(1,n+1):
            if i<=len(a2)*3 and i%3==0 and j<=len(a2)-1:
                ans[a2[j]]=i
                j+=1
            else:
                ans[a1[k]]=i
                k+=1

else:
    for i in range(1,n+1):
        if i%3==1:
            ans[a1[j]]=i
            j+=1
        elif i%3==2:
            ans[a2[k]] = i
            k += 1
        else:
            continue

    for i in range(3,n+1,3):
        if j<=len(a1)-1:
            ans[a1[j]]=i
            j+=1
        else:
            ans[a2[k]] = i
            k += 1


print(*ans)