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 defaultdict, deque


def threeSets(vs, es, d):
    sets = [3] * vs
    
    s1 = 0
    dist = [float('inf')] * vs
    sets[s1] = 1
    dist[s1] = 0
    queue = deque([s1])
    while queue:
        v1 = queue.pop()
        for v2 in d[v1]:
            if dist[v2] > dist[v1] + 1:
                dist[v2] = dist[v1] + 1
                queue.appendleft(v2)
    
    for i in range(vs):
        if dist[i] > 2:
            return [-1]
        elif dist[i] == 2:
            sets[i] = 1
        
    try:
        s2 = sets.index(3)
    except:
        return [-1]
    
    dist = [float('inf')] * vs
    sets[s2] = 2
    dist[s2] = 0
    queue = deque([s2])
    while queue:
        v1 = queue.pop()
        for v2 in d[v1]:
            if dist[v2] > dist[v1] + 1:
                dist[v2] = dist[v1] + 1
                queue.appendleft(v2)
    
    for i in range(vs):
        if dist[i] > 2:
            return [-1]
        elif dist[i] == 2:
            if sets[i] == 1:
                return [-1]
            else:
                sets[i] = 2
                
    VS = [set() for i in range(3)]
    for i in range(vs):
        g = sets[i] - 1
        VS[g].add(i)
    for V in VS:
        if not len(V):
            return [-1]
        
    for v1 in VS[0]:
        for v2 in VS[1]:
            if v2 not in d[v1]:
                return [-1]
        
        for v2 in VS[2]:
            if v2 not in d[v1]:
                return [-1]
    
    for v1 in VS[1]:
        for v2 in VS[2]:
            if v2 not in d[v1]:
                return [-1]
            
    valid_es = len(VS[0]) * len(VS[1]) + len(VS[0]) * len(VS[2]) + len(VS[1]) * len(VS[2])
    return sets if es == valid_es else [-1]       
        
    
d = defaultdict(set)

vs, es = map(int, input().split())
for _ in range(es):
    v1, v2 = map(int, input().split())
    v1 -= 1
    v2 -= 1
    d[v1].add(v2)
    d[v2].add(v1)
        
print(*threeSets(vs, es, d))
