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:
def main():
    n = int(input())
    ab = [list(map(int, input().split())) for _ in [0]*(n-1)]
    g = [[] for _ in [0]*n]
    [g[a-1].append(b-1) for a, b in ab]
    [g[b-1].append(a-1) for a, b in ab]

    root = 0  # 根
    d = [-1]*n  # 根からの距離
    d[root] = 0
    q = [root]
    cnt = 0
    while q:  # BFS
        cnt += 1
        qq = []
        while q:
            i = q.pop()
            for j in g[i]:
                if d[j] == -1:
                    d[j] = cnt
                    qq.append(j)
        q = qq
    d = [i % 2+1 for i in d]
    d2 = [0]*n
    x = d.count(1)
    y = d.count(2)
    if x > n//3 and y > n//3:
        j1 = 1
        for i in range(n):
            if d[i] == 1:
                d2[i] = j1
                j1 += 3
                cnt += 1
        j2 = 2
        for i in range(n):
            if d[i] == 2:
                d2[i] = j2
                j2 += 3
        j3 = 3
        for i in range(n):
            if d2[i] > n:
                d2[i] = j3
                j3 += 3
    elif x <= n//3:
        j3 = 3
        for i in range(n):
            if d[i] == 1:
                d2[i] = j3
                j3 += 3
        j2 = 1
        for i in range(n):
            if d[i] == 2:
                d2[i] = j2
                if j2 % 3 == 2:
                    j2 += 2
                else:
                    j2 += 1
        for i in range(n):
            if d2[i] > n:
                d2[i] = j3
                j3 += 3
    else:
        j3 = 3
        for i in range(n):
            if d[i] == 2:
                d2[i] = j3
                j3 += 3
        j2 = 1
        for i in range(n):
            if d[i] == 1:
                d2[i] = j2
                if j2 % 3 == 2:
                    j2 += 2
                else:
                    j2 += 1
        for i in range(n):
            if d2[i] > n:
                d2[i] = j3
                j3 += 3
    print(*d2)


main()