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

sys.setrecursionlimit(2 * 10**6)


def inpl():
    return list(map(int, input().split()))


def dfs(n, d):
    if depth[n] != -1:
        return
    depth[n] = d
    for nn in eAB[n]:
        dfs(nn, d + 1)
    return


N = int(input())
AB = [inpl() for i in range(N - 1)]

eAB = [[] for i in range(N)]
for a, b in AB:
    eAB[a - 1].append(b - 1)
    eAB[b - 1].append(a - 1)

depth = [-1 for i in range(N)]

dfs(0, 0)

eo = [[], []]
for i in range(N):
    eo[depth[i] % 2].append(i)

# print(eo)
ans = [-1 for i in range(N)]
if len(eo[0]) > N // 3 and len(eo[1]) > N // 3:
    for i in range(N // 3):
        ans[eo[0][i]] = 3 * i + 1
        ans[eo[1][i]] = 3 * i + 2
    t_0 = N // 3
    t_1 = N // 3
    if N % 3 > 0:
        ans[eo[0][N // 3]] = 3 * (N // 3) + 1
        t_0 += 1
    if N % 3 > 1:
        ans[eo[1][N // 3]] = 3 * (N // 3) + 2
        t_1 += 1
    # print(ans)
    for i, v in enumerate(eo[0][t_0:] + eo[1][t_1:]):
        ans[v] = 3 * (i + 1)
else:
    if len(eo[0]) <= N // 3:
        ni = 0
    else:
        ni = 1
    tmp = [True for i in range(N + 1)]
    for i in range(len(eo[ni])):
        ans[eo[ni][i]] = 3 * (i + 1)
        tmp[3 * (i + 1)] = False
    tmp = [i for i, t in enumerate(tmp) if t is True]
    j = 1
    for i in range(N):
        if ans[i] != -1:
            continue
        ans[i] = tmp[j]
        j += 1

print(" ".join(list(map(str, ans))))