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

xs = [[] for _ in range(2)]
g2 = [set() for _ in range(N+1)]
vs = set()
stack = [1]
depth = [0]*(N+1)
dfs_vs = []
vs.add(1)
while stack:
    v = stack.pop()
    xs[depth[v] % 2].append(v)
    dfs_vs.append(v)
    for u in G[v]:
        if u in vs:
            continue
        depth[u] = depth[v] + 1
        g2[v].add(u)
        stack.append(u)
        vs.add(u)

xs.sort(key=lambda x:len(x))
rs = []
if len(xs[0]) <= N//3:
    i=j=k=0
    for x in xs[0]:
        i+=1
        rs.append((x, 3*i))
    for x in xs[1]:
        if i+1 <= N//3:
            i+=1
            rs.append((x, 3*i))
        elif j <= (N-1)//3:
            rs.append((x, 3*j+1))
            j+=1
        else:
            rs.append((x, 3*k+2))
            k+=1
else:
    i=j=k=0
    for x in xs[0]:
        if j <= (N-1)//3:
            rs.append((x, 3*j+1))
            j+=1
        elif i+1 <= N//3:
            i+=1
            rs.append((x, 3*i))
    for x in xs[1]:
        if i+1 <= N//3:
            i+=1
            rs.append((x, 3*i))
        else:
            rs.append((x, 3*k+2))
            k+=1

#print(rs)
rs.sort(key=lambda x:x[0])
print(" ".join(str(x[1]) for x in rs))