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 Counter, defaultdict
from sys import stdin
raw_input = stdin.readline

n,m=map(int,raw_input().split())
d=defaultdict(list)
st=[[] for i in xrange(3)]
c=0
for i in xrange(m):
    u,v = map(int,raw_input().split())
    d[u].append(v)
    d[v].append(u)
vis=[0]*(n+1)
for i in xrange(1,n+1):
    if not vis[i]:
        if c>2:
            print -1
            exit()
        
        
        x=Counter(d[i])
        for j in xrange(1,n+1):
            
            if (not x[j]) and (not vis[j]):
                st[c].append(j)
                vis[j]=1
                for k in d[j]:
                    x[k]=1
                    
            elif (not x[j]) and vis[j]:
                print -1
                exit()
        c+=1

ans=[0]*(n+1)
val=1
c1=len(st[0])
c2=len(st[1])
c3=len(st[2])
if m!=(c1*c2)+(c2*c3)+(c3*c1):
    print -1
    exit()
for set in st:
    if not set:
        print -1
        exit()
    for j in set:
        ans[j]=val
    val+=1
for i in xrange(1,n+1):
    print ans[i],
print
