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())
from collections import defaultdict
flag = [0 for i in range(N)]
G = defaultdict(list)
for _ in range(N-1):
    a,b = map(int,input().split())
    G[a-1].append(b-1)
    G[b-1].append(a-1)

# G[v]: 頂点vに隣接する頂点list
# N: 頂点数

from collections import deque
dist = [-1]*N
que = deque([0])
dist[0] = 0
while que:
    v = que.popleft()
    d = dist[v]
    for w in G[v]:
        if dist[w] > -1:
            continue
        dist[w] = d + 1
        que.append(w)


al = [0, 0]
for d in dist:
    al[d%2] += 1
if sum(al) // al[0] >= 3:
    numbers1 = [i for i in range(1, N + 1) if i % 3 == 1]
    numbers2 = [i for i in range(1, N + 1) if i % 3 == 2]
    numbers3 = [i for i in range(1, N + 1) if i % 3 == 0]
    ansl = []
    for d in dist:
        if d % 2 == 0:
            ansl.append(numbers3.pop())
        else:
            if numbers1:
                ansl.append(numbers1.pop())
            elif numbers2:
                ansl.append(numbers2.pop())
            else:
                ansl.append(numbers3.pop())
    for a in ansl:
        print(a)
elif sum(al) // al[1] >= 3:
    numbers1 = [i for i in range(1, N + 1) if i % 3 == 1]
    numbers2 = [i for i in range(1, N + 1) if i % 3 == 2]
    numbers3 = [i for i in range(1, N + 1) if i % 3 == 0]
    ansl = []
    for d in dist:
        if d % 2 == 0:
            if numbers1:
                ansl.append(numbers1.pop())
            elif numbers2:
                ansl.append(numbers2.pop())
            else:
                ansl.append(numbers3.pop())
        else:
            ansl.append(numbers3.pop())
    for a in ansl:
        print(a)
else:
    numbers1 = [i for i in range(1, N + 1) if i % 3 == 1]
    numbers2 = [i for i in range(1, N + 1) if i % 3 == 2]
    numbers3 = [i for i in range(1, N + 1) if i % 3 == 0]
    ansl = []
    for d in dist:
        if d % 2 == 0:
            if numbers1:
                ansl.append(numbers1.pop())
            else:
                ansl.append(numbers3.pop())
        else:
            if numbers2:
                ansl.append(numbers2.pop())
            else:
                ansl.append(numbers3.pop())
    for a in ansl:
        print(a)