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 sys
input = sys.stdin.readline

n,m=map(int,input().split())
E=[set() for i in range(n+1)]
ECOUNT=[0]*(n+1)

for i in range(m):
    x,y=map(int,input().split())
    E[x].add(y)
    E[y].add(x)
    ECOUNT[x]+=1
    ECOUNT[y]+=1


Group=[i for i in range(n+1)]

def find(x):
    while Group[x] != x:
        x=Group[x]
    return x

def Union(x,y):
    if find(x) != find(y):
        Group[find(y)]=Group[find(x)]=min(find(y),find(x))

SCORE=0
for i in range(1,n+1):
    if find(i)==i:
        SCORE+=1
        if SCORE>=4:
            print(-1)
            sys.exit()
            
        for j in range(i+1,n+1):
            if j in E[i]:
                continue
            else:
                Union(i,j)

FD=[find(i) for i in range(n+1)]
if len(set(FD[1:]))!=3:
    print(-1)
    sys.exit()

for i in range(n+1):
    for j in E[i]:
        if FD[i]==FD[j]:
            print(-1)
            sys.exit()


compression_dict={a: ind for ind, a in enumerate(sorted(set(FD)))}
ANS=[compression_dict[a] for a in FD]

VFD=[0,0,0]
EFD=[0,0,0]
for i in range(1,n+1):
    VFD[ANS[i]-1]+=1
    EFD[ANS[i]-1]+=ECOUNT[i]

for i in range(3):
    VS=0
    for j in range(3):
        if i==j:
            continue
        VS+=VFD[i]*VFD[j]

    if EFD[i]==VS:
        continue
    else:
        print(-1)
        sys.exit()


print(*ANS[1:])


