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.
import sys
n, m = [int(i) for i in sys.stdin.readline().split()]

group = [0]*n

data = [set() for i in range(n)]

gi = 0
for i in range(m):
    a, b = [int(j)-1 for j in sys.stdin.readline().split()]
    data[a].add(b)
    data[b].add(a)


for i in range(n):
    if group[i] == 0:
        gi += 1
        if gi == 4:
            print(-1)
            exit()
        
        nei = data[i]
        for j in range(n):
            if j not in nei:
                group[j] = gi

cnt = [0,0,0, 0]
cnt[1] = group.count(1)
cnt[2] = group.count(2)
cnt[3] = n - cnt[1] - cnt[2]


for i in range(1, 4):
    if cnt[i] <= 0:
        print(-1)
        exit()
# print(data)
# print(cnt)
# print(group)

# dic = {}

amount = 0
for a, b in [(1,2), (2,3), (1,3)]:
    amount += cnt[a] * cnt[b]
    

# print(amount)

if amount != m:
    print(-1)
    exit()
    
for a in range(1,4):    
    
    for i in range(n):
        if group[i] == a:
            for neib in data[i]:
                c = group[neib]
                if c == a:
                    print(-1)
                    exit()
                



# for a, b in [(1,2), (2,3), (1,3)]:
#     if dic[(a,b)] != 0:
#         print(-1)
#         exit()


print(*group)
