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


def part(s, links):
    grp = [set(), set()]
    q = [(s, -1, 0)]
    while q:
        v, p, g = q.pop()
        grp[g].add(v)
        for u in links[v]:
            if u == p:
                continue
            q.append((u, v, g ^ 1))
    return grp


def solve(n, links):
    grp = part(0, links)
    can = [[], [], []]
    for i in range(1, n + 1):
        can[i % 3].append(i)
    # print(grp)
    # print(can)

    if len(grp[0]) > len(grp[1]):
        grp[0], grp[1] = grp[1], grp[0]
    if len(can[1]) > len(can[2]):
        can[1], can[2] = can[2], can[1]

    ans = [0] * n

    if len(grp[0]) <= len(can[0]):
        for i, v in enumerate(grp[0]):
            ans[v] = can[0].pop()
        can = can[0] + can[1] + can[2]
        for i, v in enumerate(grp[1]):
            ans[v] = can.pop()
        return ans

    for i, v in enumerate(grp[0]):
        if can[1]:
            ans[v] = can[1].pop()
        else:
            ans[v] = can[0].pop()
    for i, v in enumerate(grp[1]):
        if can[2]:
            ans[v] = can[2].pop()
        else:
            ans[v] = can[0].pop()
    return ans


n = int(input())
links = [set() for _ in range(n)]
for line in sys.stdin:
    a, b = map(int, line.split())
    a -= 1
    b -= 1
    links[a].add(b)
    links[b].add(a)
print(*solve(n, links))