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.
import sys

input = []

for line in sys.stdin:
	input.append(line.split())

n = int(input[0][0])
m = int(input[0][1])

if n < 3 or m < 3:
	sys.stdout.write('-1')
	exit()

edges = dict()

def is_edge(l, r):
	if l > r:
		l, r = r, l
	if l in edges:
		if r in edges[l]:
			return True
	return False

for i in range(1, len(input)):
	start = int(input[i][0])
	end = int(input[i][1])

	if start not in edges:
		edges[start] = set()
	edges[start].add(end)

c = [set(), set(), set()]

l = int(input[1][0])
r = int(input[1][1])

c[0].add(l)
c[1].add(r)

for i in range(1, n + 1):
	if i == l or i == r:
		continue

	el = is_edge(l, i)
	er = is_edge(r, i)

	if el and er:
		c[2].add(i)
	elif not el and er:
		c[0].add(i)
	elif el and not er:
		c[1].add(i)
	else:
		sys.stdout.write('-1')
		exit()

if m != (len(c[0]) * len(c[1]) + len(c[1]) * len(c[2]) + len(c[2]) * len(c[0])):
	sys.stdout.write('-1')
	exit()

if len(c[0]) == 0 or len(c[1]) == 0 or len(c[2]) == 0:
	sys.stdout.write('-1')
	exit()

for node, to_edges in edges.iteritems():
	for to in to_edges:
		if (node in c[0] and to in c[0]) or (node in c[1] and to in c[1]) or (node in c[2] and to in c[2]):
			sys.stdout.write('-1')
			exit()

for i in range(1, n + 1):
	for j in range(0, 3):
		if i in c[j]:
			sys.stdout.write(str(j + 1))
			if i != n:
				sys.stdout.write(' ')







