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.
# https://codeforces.com/contest/1228/problem/D
# all neightbor in group --> pass 1
# all neighbor not in group --> merge 0
# invalid 2
# WA
def type_(list_v, group):
    cnt_0 = 0
    cnt_1 = 0
    
    for v in list_v:
        if v in group:
            cnt_1 += 1
        else:
            cnt_0 += 1
            
    if cnt_1 == len(group):
        return 1
    
    if cnt_0 == len(list_v):
        return 0
    
    return 2
    
def is_all_type_1(ex_index, list_group, v):
    for i, group in list_group.items():
        if i == ex_index:
            continue
            
        if type_(g[v], group) != 1: 
            return False
        
    return True
    
def check(v, list_group):
    t = None
    for i, group in list_group.items():
        t  = type_(g[v], group)
        
        if t == 0 or t == 2:
            if t == 0:
                if is_all_type_1(i, list_group, v) == True:
                    group[v] = 1
                else:
                    return 2
            return t
        
    return t    
    
group = {}    
def process(g):    
    for v in g:
        if len(group) == 0:
            group[0]    = {}
            group[0][v] = 1
            continue
    
        t = check(v, group)
        
        if t == 2:
            return -1
        
        if t == 1:
            if len(group) == 3:
                return -1
            
            group[len(group)]    = {}
            group[len(group)-1][v] = 1
    return group

g = {}
n, m = map(int, input().split())

for _ in range(m):
    u, v = map(int, input().split())
    if u not in g:
        g[u] = []
    if v not in g:
        g[v] = []
        
    g[u].append(v)    
    g[v].append(u)
    
ans = process(g)

if ans == -1 or len(ans) < 3:
    print(-1)
else:
    pr  = [0] * n
    
    cnt = 0
    for k, gr in group.items():
        for v in gr:
            cnt += 1
            pr[v-1] = str(k+1)
    
    if cnt == n:
        print(' '.join(pr))
    else:
        print(-1)
# 1,2  3,4  5,6
#6 12
#1 3
#1 4
#2 3
#2 4
#1 5 
#1 6
#2 5
#2 6
#3 5
#3 6
#4 5
#4 6