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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, i, j, k, x, g, a, b;
  cin >> n >> m;
  vector<set<int> > v(n + 1);
  for (i = 0; i < m; i++) {
    cin >> a >> b;
    v[a - 1].insert(b - 1);
    v[b - 1].insert(a - 1);
  }
  vector<int> group(n + 1, -1);
  for (g = 0; g < 3; g++) {
    for (i = 0; i < n; i++) {
      if (group[i] == -1) {
        break;
      }
    }
    if (i == n + 1) {
      cout << -1 << endl;
      return 0;
    }
    group[i] = g;
    for (j = 0; j < n; j++) {
      if (i != j && v[i].find(j) == v[i].end()) {
        if (group[j] != -1) {
          cout << -1 << endl;
          return 0;
        } else {
          group[j] = g;
        }
      }
    }
  }
  vector<vector<int> > gr(3);
  for (i = 0; i < n; i++) {
    if (group[i] == -1) {
      cout << -1 << endl;
      return 0;
    } else {
      gr[group[i]].push_back(i);
    }
  }
  int found_edges = 0;
  for (i = 0; i < 2; i++) {
    for (j = i + 1; j < 3; j++) {
      for (int v1 : gr[i])
        for (int v2 : gr[j]) {
          if (v[v1].find(v2) == v[v1].end()) {
            printf("-1\n");
            return 0;
          } else
            found_edges++;
        }
    }
  }
  if (found_edges < m) {
    cout << -1 << endl;
    return 0;
  } else {
    for (i = 0; i < n; i++) {
      cout << group[i] + 1 << " ";
    }
    cout << endl;
  }
  return 0;
}
