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.
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline


class UnionFind:
    def __init__(self, N):
        self.parent = [i for i in range(N)]
        self.size = [1 for _ in range(N)]

    def find(self, x):
        if self.parent[x] == x:
            return x
        else:
            return self.find(self.parent[x])

    def union(self, x, y):
        px = self.find(x)
        py = self.find(y)
        if px == py:
            return
        if self.size[px] < self.size[py]:
            self.parent[px] = py
            self.size[py] += self.size[px]
        else:
            self.parent[py] = px
            self.size[px] += self.size[py]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def connectedNum(self, x):
        return self.size[self.find(x)]

    def component_NUM(self):
        par = set()
        for i in self.parent:
            par.add(self.find(i))
        return len(par)


N, M = map(int, input().split())
adj = [set() for _ in range(N)]
Un = UnionFind(N)
for _ in range(M):
    a, b = map(lambda x: int(x)-1, input().split())
    adj[a].add(b)
    adj[b].add(a)

added = set()
representative = set()
for i in range(N):
    if i in added:
        continue
    added.add(i)
    representative.add(i)
    for j in range(i+1, N):
        if j in added:
            continue
        if adj[i] == adj[j]:
            added.add(j)
            Un.union(i, j)
    if len(representative) > 3:
        print(-1)
        exit()
if Un.component_NUM() == 3:
    group = {}
    ans = []
    for p in range(N):
        par = Un.find(p)
        if par not in group.keys():
            group[par] = len(group)+1
        ans.append(group[par])
    print(" ".join(map(str, ans)))
else:
    print(-1)
