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.

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In the second example, it's impossible to make such vertex sets.
import sys
input = sys.stdin.readline
from collections import deque
n, m = [int(item) for item in input().split()]
ab = []
edges = [[] for _ in range(n)]

for i in range(m):
    a, b = [int(item) for item in input().split()]
    a -= 1; b -= 1
    edges[a].append(b)
    edges[b].append(a)
    ab.append((a, b))

    
groupA = [1] * n
for a, b in ab:
    if a == 0:
        groupA[b] = 0
    elif b == 0:
        groupA[a] = 0
par = None
for i, item in enumerate(groupA):
    if item == 0:
        par = i
        break
if par == None:
    print(-1)
    exit()

groupB = [1] * n
for a, b in ab:
    if a == par:
        groupB[b] = 0
    elif b == par:
        groupB[a] = 0

par = None
for i, (p, q) in enumerate(zip(groupA, groupB)):
    if p == 0 and q == 0:
        par = i
        break
if par == None:
    print(-1)
    exit()

groupC = [1] * n
for a, b in ab:
    if a == par:
        groupC[b] = 0
    elif b == par:
        groupC[a] = 0

# Check edge num
sumA = sum(groupA)
sumB = sum(groupB)
sumC = sum(groupC)
e_abc = [0, n - sumA, n - sumB, n - sumC]
edge_num = sumA * sumB + sumB * sumC + sumC * sumA 
if edge_num != m:
    print(-1)
    exit()

# Answer
setA = set()
setB = set()
setC = set()
group = [groupA, groupB, groupC]
ret = []
for i, (ga, gb, gc) in enumerate(zip(groupA, groupB, groupC)):
    total = ga + gb + gc
    if total != 1:
        print(-1)
        exit()
    if ga:
        ret.append(1)
        setA.add(i)
    elif gb:
        ret.append(2)
        setB.add(i)
    else:
        ret.append(3)
        setC.add(i)
s_ABC = [set(), setA, setB, setC]
for i, item in enumerate(ret):
    if e_abc[item] != len(edges[i]):
        print(-1)
        exit()
    for node in edges[i]:
        if node in s_ABC[item]:
            print(-1)
            exit()
print(" ".join([str(item) for item in ret]))