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():
    import sys
    input = sys.stdin.readline

    N = int(input())
    adj = [[] for _ in range(N+1)]
    for _ in range(N-1):
        a, b = map(int, input().split())
        adj[a].append(b)
        adj[b].append(a)

    from collections import deque
    que = deque()
    que.append(1)
    seen = [-1] * (N+1)
    seen[1] = 0
    par = [0] * (N+1)
    child = [[] for _ in range(N+1)]
    seq = []
    one_num = 0
    while que:
        v = que.popleft()
        seq.append(v)
        for u in adj[v]:
            if seen[u] == -1:
                seen[u] = seen[v] ^ 1
                if seen[u]:
                    one_num += 1
                par[u] = v
                child[v].append(u)
                que.append(u)
    seq.reverse()
    two_num = N - one_num

    one = list(range(1, N+1, 3))
    two = list(range(2, N+1, 3))
    three = list(range(3, N+1, 3))
    three_num = N // 3

    ans = [0] * (N+1)
    if one_num <= three_num:
        for v in range(1, N+1):
            if seen[v]:
                ans[v] = three.pop()
            else:
                if one:
                    ans[v] = one.pop()
                elif two:
                    ans[v] = two.pop()
                else:
                    ans[v] = three.pop()
    elif two_num <= three_num:
        for v in range(1, N+1):
            if not seen[v]:
                ans[v] = three.pop()
            else:
                if one:
                    ans[v] = one.pop()
                elif two:
                    ans[v] = two.pop()
                else:
                    ans[v] = three.pop()
    else:
        for v in range(1, N+1):
            if seen[v]:
                if one:
                    ans[v] = one.pop()
                else:
                    ans[v] = three.pop()
            else:
                if two:
                    ans[v] = two.pop()
                else:
                    ans[v] = three.pop()
    print(*ans[1:])


if __name__ == '__main__':
    main()