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

sr = lambda: sys.stdin.readline().rstrip()
ir = lambda: int(sr())
lr = lambda: list(map(int, sr().split()))

# 1(mod 3)のグループと2(mod 3)を先に決める
N = ir()
graph = [[] for _ in range(N+1)] # 1-indexed
for _ in range(N-1):
    a, b = lr()
    graph[a].append(b)
    graph[b].append(a)

root = 1
rank = [0] * (N+1) # 1-indexed
rank[1] = 1
parent = [0] * (N+1) # 1-indexed
stack = [root]
while stack:
    cur = stack.pop()
    r = rank[cur]
    for next in graph[cur]:
        if next == parent[cur]:
            continue
        parent[next] = cur
        rank[next] = r + 1
        stack.append(next)

counter = [0, 0]
for x in rank[1:]:
    counter[x%2] += 1
num0mod3 = N // 2
num1mod3 = (N+2) // 3
num2mod3 = (N+1) // 3
answer = [0] * (N+1)
if max(counter) >= num1mod3 + num2mod3:
    #この時ここに1mod3、2mod3を全部入れる
    if counter[0] > counter[1]:
        index = 0
    else:
        index = 1
    cur = 1
    used = set()
    for i in range(1, N+1):
        if rank[i] % 2 == index:
            answer[i] = cur
            used.add(cur)
            cur += 1
            if cur % 3 == 0:
                cur += 1
            if cur > N:
                break
    cur = 1
    for j in range(1, N+1):
        while cur in used:
            cur += 1
        if answer[j] == 0:
            answer[j] = cur
            used.add(cur)
else:
    #ここではcounterの多い方にnum1mod3を入れる
    if counter[0] > counter[1]:
        index = 0
    else:
        index = 1
    cur = 1
    used = set()
    for i in range(1, N+1):
        if rank[i] % 2 == index:
            answer[i] = cur
            used.add(cur)
            cur += 3
            if cur > N:
                break
    cur = 2
    for i in range(1, N+1):
        if rank[i] % 2 == 1-index:
            answer[i] = cur
            used.add(cur)
            cur += 3
            if cur > N:
                break
    cur = 1
    for j in range(1, N+1):
        while cur in used:
            cur += 1
        if answer[j] == 0:
            answer[j] = cur
            used.add(cur)

print(*answer[1:])