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 os
import sys
from io import BytesIO, IOBase
from collections import Counter
import math as mt

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)


def gcd(a, b):
    if a == 0:
        return b
    return gcd(b % a, a)


def lcm(a, b):
    return (a * b) / gcd(a, b)


mod = int(1e9) + 7


def power(k, n):
    if n == 0:
        return 1
    if n % 2:
        return (power(k, n - 1) * k) % mod
    t = power(k, n // 2)
    return (t * t) % mod


def totalPrimeFactors(n):
    count = 0
    if (n % 2) == 0:
        count += 1
        while (n % 2) == 0:
            n //= 2

    i = 3
    while i * i <= n:
        if (n % i) == 0:
            count += 1
            while (n % i) == 0:
                n //= i
        i += 2
    if n > 2:
        count += 1
    return count


# #MAXN = int(1e7 + 1)
# # spf = [0 for i in range(MAXN)]
#
#
# def sieve():
#     spf[1] = 1
#     for i in range(2, MAXN):
#         spf[i] = i
#     for i in range(4, MAXN, 2):
#         spf[i] = 2
#
#     for i in range(3, mt.ceil(mt.sqrt(MAXN))):
#         if (spf[i] == i):
#             for j in range(i * i, MAXN, i):
#                 if (spf[j] == j):
#                     spf[j] = i
#
#
# def getFactorization(x):
#     ret = 0
#     while (x != 1):
#         k = spf[x]
#         ret += 1
#         # ret.add(spf[x])
#         while x % k == 0:
#             x //= k
#
#     return ret


# Driver code

# precalculating Smallest Prime Factor
# sieve()

def main():
    n, m = map(int, input().split())
    d = {}
    ans = [-1 for i in range(n + 1)]
    U, V, Z = -1, -1, -1
    edges=[]
    adj=[[] for i in range(n+1)]
    for i in range(m):
        u, v = map(int, input().split())
        adj[u].append(v)
        adj[v].append(u)
        edges.append([u, v])
        if i == 0:
            ans[u] = 1
            ans[v] = 2
            U = u
            V = v
        d[(u, v)] = d[(v, u)] = 1
    f=1
    o, tw, th=1, 1, 0
    for i in range(1, n + 1):
        if (U, i) in d.keys() and (V, i) in d.keys():
            ans[i] = 3
            Z = i
            th+=1
            break
    if Z!=-1:
        for i in range(1, n + 1):
            if i != U and i != V and i != Z:
                if (i, U) in d.keys() and (i, V) in d.keys() and (i, Z) not in d.keys():
                    ans[i] = 3
                    th+=1
                elif (i, U) in d.keys() and (i, V) not in d.keys() and (i, Z) in d.keys():
                    ans[i] = 2
                    tw+=1
                elif (i, U) not in d.keys() and (i, V) in d.keys() and (i, Z) in d.keys():
                    ans[i] = 1
                    o+=1
                else:
                    f = 0
    else:
        f=0
    sums=[o, tw, th]
    k=sum(sums)
    for i in range(1, n+1):
        if len(adj[i])!=k-sums[ans[i]-1]:
            f=0
    for i in edges:
        if ans[i[0]]==-1 or ans[i[1]]==-1:
            f=0
        if ans[i[0]]==ans[i[1]]:
            f=0
    if f:
        for i in range(1, n + 1):
            print(ans[i], end=' ')
    else:
        print(-1)

    return


if __name__ == "__main__":
    main()
