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.
'''
 分组
'''
def split_group(n):
    for g in range(1, 4):
        first_node = 1
        while first_node<=n and group[first_node]!=0:
            first_node=first_node+1
        if first_node>n:
            return False
        group[first_node] = g
        group_cnt[g] = group_cnt[g]+1
        group_mem[g].append(first_node)
        #print(first_node)
        for i in range(1,n+1):
            if group[i]==0 and (i not in edge[first_node]):
                group[i] = g
                group_cnt[g] = group_cnt[g]+1
                group_mem[g].append(i)
 
'''
 检验
'''
def judge():
    if group_cnt[1]*group_cnt[2]+group_cnt[2]*group_cnt[3]+group_cnt[3]*group_cnt[1] != m:
        return False
    for g1 in range(1, 3):
        for g2 in range(g1+1, 4):
            for node1 in group_mem[g1]:
                for node2 in group_mem[g2]:
                    if node1 not in edge[node2]:
                        return False
    return True
 
edge = dict()
 
n, m = [int(c) for c in input().split()]
group = [0]*(n+1)
group_cnt = [0]*4 #三个组当前节点数
group_mem = [[],[],[],[]] #记录每个组有哪些节点
 
#print(group)
for i in range(n):
    edge[i+1] = set()
 
for i in range(m):
    u, v = [int(c) for c in input().split()]
    edge[u].add(v)
    edge[v].add(u)
 
#print(edge)
if split_group(n)==False:
    print(-1)
    
else:
    if judge()==False or m==0:
        print(-1)
    else:
        for node_group in range(1, n+1):
            print(group[node_group], end=' ')