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;
  cin >> n >> m;
  vector<unordered_set<int>> g(n);
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    g[a].insert(b);
    g[b].insert(a);
  }
  vector<int> d(n, -1);
  vector<int> cnt(3, 0);
  for (int color = 0; color < 3; color++) {
    int pos = -1;
    for (int i = 0; i < n; i++)
      if (d[i] == -1) {
        pos = i;
        break;
      }
    if (pos == -1) return cout << -1 << endl, 0;
    cnt[color]++;
    d[pos] = color;
    for (int i = 0; i < n; i++)
      if (i != pos && d[i] == -1 && !g[pos].count(i)) {
        d[i] = color;
        cnt[color]++;
      }
  }
  vector<vector<int>> groups(3);
  for (int i = 0; i < n; i++) {
    if (d[i] == -1) return cout << -1 << endl, 0;
    groups[d[i]].push_back(i);
  }
  int ff = 0;
  for (int gr1 = 0; gr1 < 3; gr1++)
    for (int gr2 = gr1 + 1; gr2 < 3; gr2++)
      for (int i : groups[gr1])
        for (int j : groups[gr2])
          if (!g[i].count(j))
            return cout << -1 << endl, 0;
          else
            ff++;
  if (ff != m) return cout << -1 << endl, 0;
  for (int i = 0; i < n; i++) cout << d[i] + 1 << ' ';
  cout << endl;
  return 0;
}
