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.Scanner;
import java.util.HashSet;
import java.util.ArrayList;

public class Main {

	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int n = s.nextInt(), m = s.nextInt();
	     ArrayList<HashSet<Integer>> edges = new ArrayList<HashSet<Integer>>();
	     for(int i = 0; i < n; i++) {
	          edges.add(new HashSet<Integer>());
	     }
	     //int[] arr = new int[n];
		for(int i = 0; i < m; i++) {
		     int u = s.nextInt() - 1;
		     int v = s.nextInt() - 1;
		     edges.get(u).add(v);
		     edges.get(v).add(u);
		}
		int[] arr = new int[n];
		for(int i = 0; i < n; i++) {
		     if(!edges.get(0).contains(i)) {
		          arr[i] = 1;
		     }
		}
		for(int idx = 2; idx <= 3; idx++) {
		     for(int j = 0; j < n; j++)
		     if(arr[j] == 0) {
		          for(int i = 0; i < n; i++) {
		               if(!edges.get(j).contains(i) && arr[i] == 0) {
		                    arr[i] = idx;
		                   // System.out.print(i + " ");
		               }
		          }
		          break;
		     }
		}
		int n1 = 0, n2 = 0, n3 = 0;
		boolean ans = true;
		for(int i = 0; i < edges.size(); i++) {
		     if(arr[i] == 1) {
		          n1++;
		     } else if(arr[i] == 2) {
		          n2++;
		     } else if(arr[i] == 3) {
		          n3++;
		     } else {
		          ans = false;
		     }
		     for(Integer j : edges.get(i)) {
		          if(arr[i] == arr[j]) {
		               ans = false;
		          }
		     }
		}
		if(n1 * n2 + n2 * n3 + n3 * n1 != m || n1 == 0 || n2 == 0 || n3 == 0) {
		     ans = false;
		}
		if(!ans) {
		     System.out.println(-1);
		     return;
		}
		StringBuffer strbf = new StringBuffer();
		for(int i = 0; i < n; i++) {
		     strbf.append(arr[i] + " ");
		}
		System.out.println(strbf.toString());
		
	}

}
