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 os
import sys
from atexit import register
from io import BytesIO
sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
sys.stdout = BytesIO()
register(lambda: os.write(1, sys.stdout.getvalue()))
input = lambda: sys.stdin.readline().rstrip('\r\n')
n,m = map(int,input().split(" "))
connected = [[]for i in range(n+1)]
edges = []
for i in range(m):
    a,b = map(int,input().split(" "))
    connected[a].append(b)
    connected[b].append(a)
    edges.append((a,b))
flag = True
labels = [0]*(n+1)
labels[1] = 1
alln = set(range(1,n+1))
if len(connected[1]) >0:
    node2 = connected[1][0]
else:
    flag = False
if flag:
    s1 = alln-set(connected[1])
    s2 = alln-set(connected[node2])
    s3 = alln-s1-s2
    n1,n2,n3 = len(s1),len(s2),len(s3)
    if n1+n2+n3!=n or min(n1,n2,n3) == 0:
        flag = False
    for i in list(s1):
        labels[i] = 1
    for i in list(s2):
        labels[i] = 2
    for i in list(s3):
        labels[i] = 3
    cnts = [0]*3
    for a,b in edges:
        if labels[a] == labels[b]:
            flag = False
            break
        cnts[labels[a]-1] += 1
        cnts[labels[b]-1] += 1
    if flag and cnts[0] == n1*(n2+n3) and cnts[1] == n2*(n1+n3) and cnts[2] == n3*(n1+n2):
        out = []
        for i in range(1,n+1):
            out.append(str(labels[i]))
        print " ".join(out)
    else:
        print -1
else:
    print -1
