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 sys import stdin, stdout, setrecursionlimit
input = stdin.readline
# import string
# characters = string.ascii_lowercase
# digits = string.digits
# setrecursionlimit(int(1e6))
# dir = [-1,0,1,0,-1]
# moves = 'NESW'
inf = float('inf')
from functools import cmp_to_key
from collections import defaultdict as dd
from collections import Counter, deque
from heapq import *
import math
from math import floor, ceil, sqrt
def geti(): return map(int, input().strip().split())
def getl(): return list(map(int, input().strip().split()))
def getis(): return map(str, input().strip().split())
def getls(): return list(map(str, input().strip().split()))
def gets(): return input().strip()
def geta(): return int(input())
def print_s(s): stdout.write(s+'\n')

def solve():
    n, m = geti()
    edges = dd(list)
    color = [0]*(n+1)
    for i in range(m):
        u, v = geti()
        edges[u].append(v)
        edges[v].append(u)
    for i in edges:
        color[i] = 1
        for node in edges[i]:
            color[node] = 2
        break
    # print(color)
    for i in edges:
        if color[i] == 0:
            color[i] = 1
            for node in edges[i]:
                if color[node] != 2:
                    return False
    # print(color)
    for i in edges:
        if color[i] == 2:
            for node in edges[i]:
                if color[node] == 2:
                    color[node] = 3
                elif color[node] == 0:
                    return False
            break
    for i in edges:
        if color[i] == 2:
            for node in edges[i]:
                if color[node] != 3 and color[node] != 1:
                    return False
    for i in edges:
        if color[i] == 3:
            for node in edges[i]:
                if color[node] == 3 or color[node] == 0:
                    return False

    # print(color)
    one, two, three = color.count(1), color.count(2), color.count(3)
    for i in edges:
        if color[i] == 1:
            temp2 = temp3 = 0
            for node in edges[i]:
                if color[node] == 2:
                    temp2 += 1
                elif color[node] == 3:
                    temp3 += 1
                else:
                    return False
            if temp2 != two or temp3 != three:
                return False
        if color[i] == 2:
            temp1 = temp3 = 0
            for node in edges[i]:
                if color[node] == 1:
                    temp1 += 1
                elif color[node] == 3:
                    temp3 += 1
                else:
                    return False
            if temp1 != one or temp3 != three:
                return False
    if color.count(0) == 1 and one and two and three:
        return color[1:]
    return False




if __name__=='__main__':
    ans = solve()
    if ans:
        print(*ans)
    else:
        print(-1)
