You have a simple undirected graph consisting of n vertices and m edges. The graph doesn't contain self-loops, there is at most one edge between a pair of vertices. The given graph can be disconnected.

Let's make a definition.

Let v_1 and v_2 be two some nonempty subsets of vertices that do not intersect. Let f(v_{1}, v_{2}) be true if and only if all the conditions are satisfied:

  1. There are no edges with both endpoints in vertex set v_1. 
  2. There are no edges with both endpoints in vertex set v_2. 
  3. For every two vertices x and y such that x is in v_1 and y is in v_2, there is an edge between x and y. 



Create three vertex sets (v_{1}, v_{2}, v_{3}) which satisfy the conditions below;

  1. All vertex sets should not be empty. 
  2. Each vertex should be assigned to only one vertex set. 
  3. f(v_{1}, v_{2}), f(v_{2}, v_{3}), f(v_{3}, v_{1}) are all true. 



Is it possible to create such three vertex sets? If it's possible, print matching vertex set for each vertex.

Input

The first line contains two integers n and m (3 ≤ n ≤ 10^{5}, 0 ≤ m ≤ min(3 ⋅ 10^{5}, (n(n-1))/(2))) — the number of vertices and edges in the graph.

The i-th of the next m lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. The graph doesn't contain self-loops, there is at most one edge between a pair of vertices. The given graph can be disconnected.

Output

If the answer exists, print n integers. i-th integer means the vertex set number (from 1 to 3) of i-th vertex. Otherwise, print -1.

If there are multiple answers, print any.

Examples

Input


6 11
1 2
1 3
1 4
1 5
1 6
2 4
2 5
2 6
3 4
3 5
3 6


Output


1 2 2 3 3 3 

Input


4 6
1 2
1 3
1 4
2 3
2 4
3 4


Output


-1

Note

In the first example, if v_{1} = \{ 1 \}, v_{2} = \{ 2, 3 \}, and v_{3} = \{ 4, 5, 6 \} then vertex sets will satisfy all conditions. But you can assign vertices to vertex sets in a different way; Other answers like "2 3 3 1 1 1" will be accepted as well.

<image>

In the second example, it's impossible to make such vertex sets.
from collections import defaultdict

n, m = map(int, raw_input().split(' '))
edges = defaultdict(dict)
for i in range(m):
    a, b = map(int, raw_input().split(' '))
    edges[a - 1][b - 1] = True
    edges[b - 1][a - 1] = True

colors = [-1] * n


def get_vertex_set(index):
    x = 0
    for i in xrange(n):
        if colors[i] == -1:
            x = i
            break
    else:
        return []
    colors[x] = index
    s = [x]
    for i in xrange(n):
        if x != i and colors[i] == -1 and not edges[x].get(i):
            colors[i] = index
            s.append(i)
    return s


def is_good(s1, s2):
    for x in s1:
        for y in s2:
            if not edges[x].get(y):
                return False
    return True

s1 = get_vertex_set(1)
s2 = get_vertex_set(2)
s3 = get_vertex_set(3)
s4 = get_vertex_set(4)
count1 = len(s1)
count2 = len(s2)
count3 = len(s3)
count4 = len(s4)

if count4 != 0:
    print -1
elif count3 == 0:
    print -1
elif count1 * count2 + count2 * count3 + count1 * count3 != m:
    print -1
elif is_good(s1, s2) and \
        is_good(s1, s3) and \
        is_good(s2, s3):
    print ' '.join(map(str, colors))
else:
    print -1