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,deque
import sys,heapq,bisect,math,itertools,string,queue,copy,time
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_str(): return list(sys.stdin.readline().split())

N,M = inpl()

if M == 0:
    print(-1)
    exit()

cnts = [0]*N
lines = defaultdict(set)
for _ in range(M):
    a,b = inpl()
    a-=1 ; b-=1
    cnts[a] += 1
    cnts[b] += 1
    lines[a].add(b)
    lines[b].add(a)

setc = set(cnts)
L = len(setc)
ans = [-1]*N
if L == 1:
    ans[a] = 1
    ans[b] = 2
    for i in range(N):
        if ans[i] != 0:
            if i not in lines[a]:
                ans[i] = 1
            elif i not in lines[b]:
                ans[i] = 2
            else:
                ans[i] = 3
elif L == 2:
    c1,c2 = list(setc)
    if c1*2 + c2 == N*2:
        i1 = cnts.index(c1)
        i2 = cnts.index(c2)
        for i,c in enumerate(cnts):
            if c == c2:
                ans[i] = 2
            else:
                if i not in lines[i1]:
                    ans[i] = 1
                else:
                    ans[i] = 3
    elif c1 + c2*2 == N*2:
        i1 = cnts.index(c1)
        i2 = cnts.index(c2)
        for i,c in enumerate(cnts):
            if c == c1:
                ans[i] = 1
            else:
                if i not in lines[i2]:
                    ans[i] = 2
                else:
                    ans[i] = 3
    else:
        print(-1)
        exit()
elif L == 3:
    c1,c2,c3 = list(setc)
    for i,c in enumerate(cnts):
        if c == c1:
            ans[i] = 1
        elif c == c2:
            ans[i] = 2
        elif c == c3:
            ans[i] = 3
else:
    print(-1)
    exit()


if len(set(ans)) != 3:
    print(-1)
    exit()

for s in range(N):
    for t in lines[s]:
        if ans[s] == ans[t]:
            print(-1)
            exit()

print(' '.join(map(str,ans)))
