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:
from collections import deque

def bfs():
    q = deque()
    q.append(1)
    dist = [-1] * (n + 1)
    dist[1] = 0
    while q:
        p = q.popleft()
        for i in G[p]:
            if dist[i] == -1:
                dist[i] = dist[p] + 1
                q.append(i)
    return dist

n = int(input())
G = [[] for _ in range(n + 1)]
for _ in range(n - 1):
    a, b = map(int, input().split())
    G[a].append(b)
    G[b].append(a)
dist = bfs()
c1, c2 = 0, 0
group1, group2 = [], []
for i in range(1, n + 1):
    if dist[i] % 2 == 1:
        c1 += 1
        group1.append(i)
    else:
        c2 += 1
        group2.append(i)
m = n // 3
ans = [0] * (n + 1)
if m >= min(c1, c2):
    l = [3 * i + 3 for i in range(m)] + [3 * (i // 2) + (i % 2) + 1 for i in range(n - m)]
    j = 0
    if c1 < c2:
        for i in group1:
            ans[i] = l[j]
            j += 1
        for i in group2:
            ans[i] = l[j]
            j += 1
    else:
        for i in group2:
            ans[i] = l[j]
            j += 1
        for i in group1:
            ans[i] = l[j]
            j += 1
else:
    j = 1
    for i in range(c1):
        if i <= (n - 1) // 3:
            ans[group1[i]] = 3 * i + 1
        else:
            ans[group1[i]] = 3 * j
            j += 1
    for i in range(c2):
        if i <= (n - 2) // 3:
            ans[group2[i]] = 3 * i + 2
        else:
            ans[group2[i]] = 3 * j
            j += 1
ans.pop(0)
print(*ans)