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 java.util.*;
import java.io.*;

public class Main {
	public static void main(String args[]) {new Main().run();}

	FastReader in = new FastReader();
	PrintWriter out = new PrintWriter(System.out);
	void run(){
		if(!work()) {
			out.println(-1);
		}
		out.flush();
	}
	long mod=1000000007;
	long gcd(long a,long b) {
		return b==0?a:gcd(b,a%b);
	}
	ArrayList<Integer>[] graph;
	int[] color;
	boolean work() {
		int n=in.nextInt();
		int m=in.nextInt();
		graph=(ArrayList<Integer>[])new ArrayList[n];
		for(int i=0;i<n;i++)graph[i]=new ArrayList<>();
		int[] degree=new int[n];
		int[] C=new int[3];
		color=new int[n];
		Arrays.fill(color, 1);
		int one=0,two=0;
		for(int i=0;i<m;i++) {
			int n1=in.nextInt()-1;
			int n2=in.nextInt()-1;
			graph[n1].add(n2);
			graph[n2].add(n1);
			degree[n1]++;
			degree[n2]++;
			if(i==0) {
				color[n1]=1;
				color[n2]=2;
				one=n1;
				two=n2;
			}
		}
		//1.染色
		for(int nn:graph[one]) {
			color[nn]=2;
		}
		for(int nn:graph[two]) {
			if(color[nn]==2)color[nn]=3;
		}
		for(int i=0;i<n;i++) {
			C[color[i]-1]++;
		}
		if(C[2]==0)return false;
		//2.检查相邻是否颜色相同，和是否断开
		int[] count=new int[1];
		if(!dfs(0,new boolean[n],count)||count[0]!=n) {
			return false;
		}
		//3。检测入度
		for(int i=0;i<n;i++) {
			if(degree[i]!=n-C[color[i]-1]) {
				return false;
			}
		}
		for(int i=0;i<n;i++)out.print(color[i]+" ");
		return true;
	}
	private boolean dfs(int node , boolean[] vis, int[] count) {
		count[0]++;
		vis[node]=true;
		for(int nn:graph[node]) {
			if(color[nn]==color[node])return false;
			if(!vis[nn]) {
				if(!dfs(nn,vis,count))return false;
			}
		}
		return true;
	}
	
}



class FastReader
{
	BufferedReader br;
	StringTokenizer st;

	public FastReader()
	{
		br=new BufferedReader(new InputStreamReader(System.in));
	}

	public String next() 
	{
		if(st==null || !st.hasMoreElements())
		{
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}

	public int nextInt() 
	{
		return Integer.parseInt(next());
	}

	public long nextLong()
	{
		return Long.parseLong(next());
	}
}