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.
# -*- coding:utf-8 -*-

"""

created by shuangquan.huang at 1/16/20

"""

import collections
import time
import os
import sys
import bisect
import heapq
from typing import List


def solve(N, M, g):
    if M < N - 1:
        print(-1)
        return
    
    mark = [1 for _ in range(N+1)]
    mark[1] = 1
    mark[0] = 0
    
    b = 0
    for v in g[1]:
        mark[v] = 2
        b = v

    for v in g[b]:
        if mark[v] == 2:
            mark[v] = 3
            
    na, nb, nc = mark.count(1), mark.count(2), mark.count(3)
    if M != na * nb + nb * nc + na * nc or any([n == 0 for n in [na, nb, nc]]):
        print(-1)
        return
    
    ns = [0, na, nb, nc]
    
    for u in range(1, N+1):
        v = g[u]
        if len(v) != N - ns[mark[u]]:
            print(-1)
            return
        wc = collections.Counter([mark[x] for x in v])
        for k, c in wc.items():
            if c != ns[k]:
                print(-1)
                return
            
    print(' '.join(map(str, mark[1:])))

    
N, M = map(int, input().split())
g = [[] for _ in range(N+1)]
for i in range(M):
    u, v = map(int, input().split())
    g[u].append(v)
    g[v].append(u)
    
solve(N, M, g)