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())
g=[set() for i in range(n)]
cnt_e=[0]*n
for i in range(m):
    u,v=map(int,input().split())
    g[u-1].add(v-1)
    g[v-1].add(u-1)
    cnt_e[u-1]+=1
    cnt_e[v-1]+=1
group=[i for i in range(n)]
def find(x):
    while group[x]!=x:
        x=group[x]
    return x
def union(x,y):
    if find(x)!=find(y):
        group[find(x)]=group[find(y)]=min(find(x),find(y))
score=0
for i in range(n):
    if find(i)==i:
        score+=1
        if score>=6:
            print(-1)
            exit()
        for j in range(i+1,n):
            if not j in g[i]:
                union(i,j)
fd=[find(i) for i in range(n)]
if len(set(fd))!=3:
    print(-1)
    exit()
for i in range(n):
    for j in g[i]:
        if fd[i]==fd[j]:
            print(-1)
            exit()
num=list(set(fd))
ans=[0]*n
cnt_v=[0]*3
cnt_zisu=[0]*3
for i in range(n):
    if fd[i]==num[0]:
        ans[i]=1
        cnt_v[0]+=1
        cnt_zisu[0]+=cnt_e[i]
    elif fd[i]==num[1]:
        ans[i]=2
        cnt_v[1]+=1
        cnt_zisu[1]+=cnt_e[i]
    else:
        ans[i]=3
        cnt_v[2]+=1
        cnt_zisu[2]+=cnt_e[i]
flag=True
flag&=((cnt_v[0]*cnt_v[1]+cnt_v[0]*cnt_v[2])==cnt_zisu[0])
flag&=((cnt_v[0]*cnt_v[1]+cnt_v[1]*cnt_v[2])==cnt_zisu[1])
flag&=((cnt_v[2]*cnt_v[1]+cnt_v[0]*cnt_v[2])==cnt_zisu[2])
if flag:
    print(*ans)
else:
    print(-1)