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
def main():
    input = sys.stdin.readline
    N = int(input())
    G = [[] for _ in range(N)]
    for _ in range(N-1):
        a, b = map(int, input().split())
        a, b = a-1, b-1
        G[a].append(b)
        G[b].append(a)
    
    red, black = list(), list()
    visited = [False] * N
    stack = [(0, True)]
    visited[0] = True
    while stack:
        v, is_red = stack.pop()
        if is_red: red.append(v)
        else: black.append(v)
        for to in G[v]:
            if visited[to]: continue
            visited[to] = True
            stack.append((to, not is_red))
    if len(red) < len(black): red, black = black, red

    ans = [0] * N
    if len(black) <= N//3:
        n3 = 3
        for v in black: ans[v], n3 = n3, n3 + 3
        n = 1
        for v in red:
            if n3 <= N: ans[v], n3 = n3, n3 + 3
            else:
                ans[v], n = n, n + 1
                if n%3 == 0: n += 1
    else:
        n1, n2, n3 = 1, 2, 3
        for v in black:
            if n1 <= N: ans[v], n1 = n1, n1 + 3
            else: ans[v], n3 = n3, n3 + 3
        for v in red:
            if n2 <= N: ans[v], n2 = n2, n2 + 3
            else: ans[v], n3 = n3, n3 + 3
    print(*ans)

if __name__ == '__main__':
    main()