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.
"""
NTC here
"""
from sys import setcheckinterval, stdin, setrecursionlimit
setcheckinterval(1000)
setrecursionlimit(10**7)
 
# print("Case #{}: {} {}".format(i, n + m, n * m))
 
 
def iin(): return int(stdin.readline())
 
 
def lin(): return list(map(int, stdin.readline().split()))

def BFS(s, adj):
    parent = {s: None}
    color = [-1]*(len(adj))
    color[s]+=1
    u = [s]
    while u:  # runs till u is []
        nextu = []
        for i in u:
            for v in adj[i]:
                if v not in parent:
                    color[v]=(color[i]+1)%3
                    parent[v] = i
                    nextu.append(v)
                else:
                    if color[v]==color[i]:
                        color[v]=(color[v]+1)%3
        u = nextu.copy()
    for i in range(len(adj)):
        color[i]+=1
    return color



n,m=lin()
adj=[[] for i in range(n)]
for _ in range(m):
    i,j=lin()
    adj[i-1].append(j-1)
    adj[j-1].append(i-1)

sol=BFS(0, adj)
c1=[0,0,0] #count of 1,2,3
for i in sol:
    if i==0:
        print(-1)
        exit()
    c1[i-1]+=1
ch=[[0 for i in range(3)] for j in range(3)] #12,13,23

for v in range(n):
    for u in adj[v]:
        ch[sol[v]-1][sol[u]-1]+=1
#        ch[sol[u]-1][sol[v]-1]+=1
        if sol[v]==sol[u]:
            print(-1)
            exit()
a12=ch[0][1]+ch[1][0]
a13=ch[0][2]+ch[2][0]
a23=ch[1][2]+ch[2][1]
#print(a12,a13,a23,c1,sol)
if a12==2*c1[0]*c1[1] and a13==2*c1[0]*c1[2] and a23==2*c1[1]*c1[2]:
    print(*sol)
else:
    print(-1)