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
input = sys.stdin.readline

def r1():
    return int(input())
def r2():
    return map(int, input().split())

n, m = r2()
d = {}
for i in range(1, n + 1):
    d[i] = set()
for i in range(m):
    x, y = r2()
    d[x].add(y)
    d[y].add(x)
    
bc = d.get(1, [])
if len(bc) == 0:
    print(-1)
    exit()
    
ac = d.get(next(iter(bc)), [])
if len(ac) == 0:
    print(-1)
    exit()

c = bc & ac
a = ac - c
b = bc - c
ab = a | b

na = len(a)
nb = len(b)
nc = len(c)
if na + nb + nc != n or not na or not nb or not nc:
    print(-1)
else:
    for x in a:
        if d.get(x, []) != bc:
            print(-1)
            exit()
    for x in b:
        if d.get(x, []) != ac:
            print(-1)
            exit()
    for x in c:
        if d.get(x, []) != ab:
            print(-1)
            exit()
    r = ['' for i in range(n)]
    for x in a:
        r[x - 1] = '1'
    for x in b:
        r[x - 1] = '2'
    for x in c:
        r[x - 1] = '3'
    print(" ". join(r))
