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.ArrayList;
import java.util.List;
import java.util.Scanner;

public class cf1228D {
    static class Graph {
        private int n, m;
        private List<Integer>[] adj;
        public Graph(int n, int m) {
            this.n = n;
            this.m = m;
            adj = new List[n];
            for(int i = 0; i < n; i++) {
                adj[i] = new ArrayList<>();
            }
        }
        public int getN() {
            return n;
        }
        public int getM() {
            return m;
        }
        public void addEdge(int u, int v) {
            adj[u].add(v);
        }
        public void solve() {
            boolean[] vis = new boolean[n];
            int[] ans = new int[n];
            int cnt = 0;
            for(int i = 0; i < n; i++) {
                if(!vis[i]) {
                    if(cnt++ == 3) {
                        System.out.println(-1);
                        return;
                    }
                    boolean[] curvis = new boolean[n];
                    for(int v : adj[i]) {
                        curvis[v] = true;
                    }
                    for(int j = 0; j < n; j++) {
                        if(!curvis[j]) {
                            vis[j] = true;
                            ans[j] = cnt;
                            if(adj[j].size() != adj[i].size()) {
                                System.out.println(-1);
                                return;
                            }
                            for(int v : adj[j]) {
                                if(!curvis[v]) {
                                    System.out.println(-1);
                                    return;
                                }
                            }
                        }
                    }
                }
            }
            if(cnt != 3) {
                System.out.println(-1);
                return;
            }
            for(int i = 0; i < n; i++) {
                System.out.print(ans[i] + " ");
            }
            System.out.println();
        }

    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        Graph graph = new Graph(in.nextInt(), in.nextInt());
        for(int i = 0; i < graph.getM(); i++) {
            int u = in.nextInt() - 1;
            int v = in.nextInt() - 1;
            graph.addEdge(u, v);
            graph.addEdge(v, u);
        }
        graph.solve();
    }
}
