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
s1 = {1}
s2 = set()
s3 = set()
def checkbi(hash,node):

    queue = [node]

    while queue!=[]:
        z = queue.pop(0)
        # print(s1,s2,s3,z)
        if z in s1:
            yo = 1
        if z in s2:

            yo = 0
        if z in s3:
            yo = 3
        for i in hash[z]:


            if i not in s1 and i not in s2 and i not in s3:
                if yo == 1:
                    s2.add(i)

                else:
                    s1.add(i)
                bool[i] = True
                queue.append(i)
            if i in s3 and (yo == 1 or yo == 0):
                continue
            elif i in s3 and yo == 3:
              return -1
            elif (i in s1 and yo == 1) or (i in s2 and yo == 0):
              if i not in s3 :
                if i in s1:
                    s1.remove(i)
                else:
                    # if z == 4:
                    #   if z in s2:
                    #       print(yo)
                    #   print(i,yo)
                    s2.remove(i)
                bool[i] = True
                queue.append(i)

                s3.add(i)
              else:
                  return -1






    return 1


hash = defaultdict(set)

n,m = map(int,input().split())

for i in range(m):
    a,b = map(int,input().split())
    hash[a].add(b)
    hash[b].add(a)
bool = defaultdict(bool)
for i in range(1,n+1):
  if bool[i] == False:
    z = checkbi(hash,1)
    if z == -1:
          print(-1)
          exit()
if len(s1)+len(s2)+len(s3)!=n:
    print(-1)
    exit()
for i in s3:
    z = hash[i]
    if hash[i]&s1!=s1 or hash[i]&s2!=s2:
        print(-1)
        exit()
for i in s2:
    z = hash[i]
    if hash[i]&s1!=s1 or hash[i]&s3!=s3:
        print(-1)
        exit()
for i in s1:
    z = hash[i]
    if hash[i]&s2!=s2 or hash[i]&s3!=s3:
        print(-1)
        exit()

ans = []
if s1 == set():
    print(-1)
    exit()

if s2 == set():
    print(-1)
    exit()

if s3 == set():
    print(-1)
    exit()


for i in range(1,n+1):
    if i in s1:
        ans.append(1)
    elif i in s2:
        ans.append(2)
    else:
        ans.append(3)

print(*ans)