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.util.function.Function;
import java.util.stream.Collectors;

public class Solution {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        int n = scanner.nextInt();
        int m = scanner.nextInt();

        Map<Integer, Set<Integer>> graph = new HashMap<>();

        for (int i = 0; i < m; i++) {
            int v1 = scanner.nextInt() - 1;
            int v2 = scanner.nextInt() - 1;

            graph.compute(v1, (k, v) -> {
                Set<Integer> res = v == null ? new HashSet<>() : v;
                res.add(v2);
                return res;
            });

            graph.compute(v2, (k, v) -> {
                Set<Integer> res = v == null ? new HashSet<>() : v;
                res.add(v1);
                return res;
            });
        }

        try {
            System.out.println(solve(n, graph));
        } catch (Exception e) {
            System.out.println(-1);
        }
    }

    public static String solve(int n, Map<Integer, Set<Integer>> graph) {
        int[] colors = new int[n];
        int first = findFirstNotColored(colors);
        Set<Integer> adjacent = graph.getOrDefault(first, Collections.emptySet());
        color(1, colors, adjacent);

        int second = findFirstNotColored(colors);
        adjacent = graph.getOrDefault(second, Collections.emptySet());
        color(2, colors, adjacent);

        int third = findFirstNotColored(colors);
        adjacent = graph.getOrDefault(third, Collections.emptySet());
        color(3, colors, adjacent);

        Map<Integer, Long> colorsDistribution = Arrays.stream(colors)
                .boxed()
                .collect(
                        Collectors.groupingBy(Function.identity(),
                                Collectors.counting()
                        )
                );

        Set<Integer> visited = new HashSet<>();
        dfs(0, visited, graph, colors, colorsDistribution);

        if (visited.size() != n) {
            throw new RuntimeException();
        }

        return Arrays.stream(colors)
                .mapToObj(Integer::toString)
                .collect(Collectors.joining(" "));
    }

    private static void dfs(int v, Set<Integer> visited, Map<Integer, Set<Integer>> graph, int[] colors, Map<Integer, Long> colorsDistribution) {
        visited.add(v);
        int c = colors[v];

        if (colors.length - colorsDistribution.getOrDefault(c, 0L) != graph.getOrDefault(v, Collections.emptySet()).size()) {
            throw new RuntimeException();
        }

        for (int next: graph.getOrDefault(v, Collections.emptySet())) {
            if (colors[next] == c || colors[next] == 0) {
                throw new RuntimeException();
            }
            if (!visited.contains(next)) {
                dfs(next, visited, graph, colors, colorsDistribution);
            }
        }
    }

    private static int findFirstNotColored(int[] colors) {
        for (int i = 0; i < colors.length; i++) {
            if (colors[i] == 0) {
                return i;
            }
        }
        throw new RuntimeException();
    }

    private static void color(int color, int[] colors, Set<Integer> adjacent) {
        for (int i = 0; i < colors.length; i++) {
            if (!adjacent.contains(i)) {
                if (colors[i] != 0) {
                    throw new RuntimeException();
                }
                colors[i] = color;
            }
        }
    }
}
