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())
AB = [list(map(int,input().split())) for _ in [0]*(N-1)]

def dist_bfs(N,E,start):
    d = [-1]*N
    d[start] = 0
    q = [start]
    while q:
        qq = []
        for i in q:
            di = d[i]
            for j in E[i]:
                if d[j]!=-1:continue
                d[j] = di+1
                q.append(j)
        q = qq
    return d

E = [[] for _ in [0]*N]
for a,b in AB:
    a -= 1
    b -= 1
    E[a].append(b)
    E[b].append(a)


D = dist_bfs(N,E,0)

e = []
o = []
cnt = 0
for i,d in enumerate(D):
    if d%2==0:
        e.append(i)
        cnt += 1
    else:
        o.append(i)
if cnt > N-cnt:
    e,o = o,e
    cnt = N-cnt
ans = [0]*N
if cnt <= N//3:
    done = set()
    j = 3
    for i in e:
        ans[i] = j
        done.add(j)
        j += 3
    j = 1
    for i in o:
        ans[i] = j
        j += 1
        while j in done:
            j += 1
else:
    done = set()
    j = 1
    for i in e:
        ans[i] = j
        done.add(j)
        j += 3
        if j > N: j = 3
    j = 2
    for i in o:
        ans[i] = j
        j += 1
        while j in done:
            j += 1
print(*ans)