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
sys.setrecursionlimit(10**9)

N = int(input())
ab = [list(map(int,input().split())) for _ in range(N-1)]

#%%
# 1-indexed
adj_list = [[] for _ in range(N+1)]

for a,b in ab:
    adj_list[a].append(b)
    adj_list[b].append(a)

# DFSで各頂点を1，2に割り当てる．

group = [1]*(N+1)
group[0] = -1

def dfs(v,par,grp):
    # 頂点vからpar以外の子へ遷移
    # grpで頂点vのグループを持つ
    
    # 1 xor 3 = 2
    nxt_grp = grp^3
    
    for child in adj_list[v]:
        if child==par:
            continue
        group[child] = nxt_grp
        dfs(child,v,nxt_grp)
    
    return 

dfs(1,-1,1)

ans = [0]*(N+1)
used = [False]*(N+1)


group_num = [0]*2
for g in group[1:]:
    group_num[g-1] += 1
    
    
    
if min(group_num) <= N//3:
    # どちらかのグループの要素数がN//3以下のとき，そのグループに0(mod3)を割り当てる
    if min(group_num) == group_num[0]:
        minidx = 1
    else:
        minidx = 2
    
    idx = 3
    idx_n = 1
    for i,g in enumerate(group):
        if i==0:continue
        if g==minidx:
            ans[i] = idx            
            used[idx] = True
            idx += 3

    for i,g in enumerate(group):
        if i==0:continue
        if ans[i]!=0:continue
        while used[idx_n]:
            idx_n += 1
        ans[i] = idx_n
        used[idx_n] = True
        idx_n += 1
    
    print(*ans[1:])

else:
    # 順番に1,2を割り当て，Nを超えたら0を割り当て
    idx_1 = 1
    idx_2 = 2
    for i,g in enumerate(group):
        if i==0:continue
        if g==1:
            if idx_1>N:continue
            ans[i] = idx_1
            used[idx_1] = True
            idx_1 += 3
        else:
            if idx_2>N:continue
            ans[i] = idx_2
            used[idx_2] = True
            idx_2 += 3

    idx = 3
    for i in range(1,N+1):
        if ans[i]==0:
            while used[idx]:
                idx += 3
            ans[i] = idx
            used[idx] = True

    print(*ans[1:])