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 D {

	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		int m = Integer.parseInt(st.nextToken());
		ArrayList<Integer>[] g = new ArrayList[n];
		for(int i = 0; i < n; i++) {
			g[i] = new ArrayList<>();
		}
		for(int i = 0; i < m; i++) {
			st = new StringTokenizer(br.readLine());
			int u = Integer.parseInt(st.nextToken())-1;
			int v = Integer.parseInt(st.nextToken())-1;
			g[u].add(v);
			g[v].add(u);
		}
		int[] color = new int[n];
		color[0] = 1;
		for(int v: g[0]) {
			color[v] = -1;
		}
		for(int i = 0; i < n; i++) {
			if(color[i] == 0) color[i] = 1;
		}
		
		for(int i = 0; i < n; i++) {
			if(color[i] == -1) {
				color[i] = 2;
				for(int v: g[i]) {
					if(color[v] == -1) {
						color[v] = 3;
					}
				}
				for(int u = 0; u < n; u++) {
					if(color[u] == -1) {
						color[u] = 2;
					}
				}
				break;
			}
		}
		
		for(int u = 0; u < n; u++) {
			for(int v: g[u]) {
				if(color[u] == color[v]) {
					System.out.println(-1); return;
				}
			}
		}
		long[] cs = new long[3];
		for(int i = 0; i < n; i++) {
			cs[color[i]-1]++;
		}
		if(cs[0] == 0 || cs[1] == 0 || cs[2] == 0) {
			System.out.println(-1); return;
		}
		long needed = cs[0]*cs[1] + cs[1]*cs[2] + cs[0]*cs[2];
		if(needed != m) {
			System.out.println(-1); return;
		}
		else {
			StringBuilder sb = new StringBuilder();
			for(int i = 0; i < n; i++) {
				sb.append(color[i]+" ");
			}
			System.out.println(sb.toString().trim());
		}
		
	}
}
