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
input = sys.stdin.readline

N = int(input())
G = [[] for _ in range(N)]
for _ in range(N-1):
    a, b = map(lambda x: int(x)-1, input().split())
    G[a].append(b)
    G[b].append(a)
group = [-1] * N
count = [0] * 2
stack = []
group[0] = 0
count[0] = 1
stack.append(0)
while stack:
    u = stack.pop()
    for v in G[u]:
        if group[v] == -1:
            group[v] = 1 - group[u]
            count[group[v]] += 1
            stack.append(v)
zeros = list(range(3, N+1, 3))
ones = list(range(1, N+1, 3))
twos = list(range(2, N+1, 3))
ans = [0] * N
for g in range(2):
    if count[g] <= N//3:
        for i in range(N):
            if group[i] == g:
                ans[i] = zeros.pop()
            else:
                if len(twos) > 0:
                    ans[i] = twos.pop()
                elif len(ones) > 0:
                    ans[i] = ones.pop()
                else:
                    ans[i] = zeros.pop()
        print(' '.join(map(str, ans)))
        exit()
for i in range(N):
    if group[i] == 0 and len(ones) > 0:
        ans[i] = ones.pop()
    elif group[i] == 1 and len(twos) > 0:
        ans[i] = twos.pop()
    else:
        ans[i] = zeros.pop()
print(' '.join(map(str, ans)))