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 $m$ ($3 \le n \le 10^{5}$, $0 \le m \le \text{min}(3 \cdot 10^{5}, \frac{n(n-1)}{2})$) — the number of vertices and edges in the graph.

The $i$-th of the next $m$ lines contains two integers $a_{i}$ and $b_{i}$ ($1 \le a_{i} \lt b_{i} \le n$) — 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.

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.

In the second example, it's impossible to make such vertex sets.