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())

node = [[] for _ in range(N)]

for _ in range(N-1):
    a, b = map(int, input().split())
    node[a-1].append(b-1)
    node[b-1].append(a-1)

# 頂点1からの距離が奇数:1、偶数:0
group = [False] * N
# 親ノードを格納する
parents = list(range(N))
stack = [0]

# ノードの選択
while stack:
    p = stack.pop()
    # 子ノードの選択
    for c in node[p]:
        # 親ノードも選択されるので、その場合は飛ばす
        if c != parents[p]:
            # 子ノードの親を自分にする
            parents[c] = p
            # 子ノードのグループを自分と逆に設定する
            group[c] = not group[p]
            stack.append(c)

p = [0] * N
three = 3
if group.count(True) <= N // 3:
    other = 1
    # Trueのところには３の倍数を、Falseにはそれ以外を
    for i in range(N):
        if not group[i] and other <= N:
            p[i] = other
            other += other % 3
        else:
            p[i] = three
            three += 3
elif group.count(False) <= N // 3:
    other = 1
    # Trueのところには３の倍数を、Falseにはそれ以外を
    for i in range(N):
        if group[i] and other <= N:
            p[i] = other
            other += other % 3
        else:
            p[i] = three
            three += 3

else:
    one, two = 1, 2
    # Trueのところには３の倍数を、Falseにはそれ以外を
    for i in range(N):
        if not group[i] and one <= N:
            p[i] = one
            one += 3
        elif group[i] and two <= N:
            p[i] = two
            two += 3
        else:
            p[i] = three
            three += 3
print(" ".join(map(str, p)))