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 main():

    N = int(input())
    g = {}
    for i in range(N):
        g[i] = set()

    for i in range(N - 1):
        a, b = map(int, input().split())
        g[a - 1].add(b - 1)
        g[b - 1].add(a - 1)

    # BFS
    dist = [-1] * N     # [-1:initial, 0:even, 1:odd]
    dist[0] = 0
    n_even = 0
    que = deque([0])
    while len(que):
        v = que[0]
        if dist[v] % 2 == 0:
            n_even += 1
        que.popleft()
        for n in g[v]:
            if dist[n] == -1:
                dist[n] = (dist[v] + 1) % 2
                que.append(n)

    ans = [-1] * N
    if n_even <= int(N / 3):
        n3, n3q, n3r = 1, 0, 1
        for i in range(N):
            if dist[i] == 0:
                ans[i] = n3 * 3
                n3 += 1
            else:
                if n3q * 3 + n3r <= N:
                    ans[i] = n3q * 3 + n3r
                    if n3r == 1:
                        n3r = 2
                    else:
                        n3q += 1
                        n3r = 1
                else:
                    ans[i] = n3 * 3
                    n3 += 1
    elif n_even > int(N * 2 / 3):
        n3, n3q, n3r = 1, 0, 1
        for i in range(N):
            if dist[i] == 1:
                ans[i] = n3 * 3
                n3 += 1
            else:
                if n3q * 3 + n3r <= N:
                    ans[i] = n3q * 3 + n3r
                    if n3r == 1:
                        n3r = 2
                    else:
                        n3q += 1
                        n3r = 1
                else:
                    ans[i] = n3 * 3
                    n3 += 1
    else:
        n3, n31, n32 = 1, 0, 0
        for i in range(N):
            if dist[i] == 0:
                if n31 * 3 + 1 <= N:
                    ans[i] = n31 * 3 + 1
                    n31 += 1
                else:
                    ans[i] = n3 * 3
                    n3 += 1
            else:
                if n32 * 3 + 2 <= N:
                    ans[i] = n32 * 3 + 2
                    n32 += 1
                else:
                    ans[i] = n3 * 3
                    n3 += 1

    print(' '.join(map(str, ans)))


if __name__ == '__main__':
    main()