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;
template <typename T>
void _dbg(const char* _s, T _h) {
  cerr << _s << " = " << _h << "\n";
}
template <typename T, typename... Ts>
void _dbg(const char* _s, T _h, Ts... _t) {
  int _b = 0;
  while (((_b += *_s == '(') -= *_s == ')') != 0 || *_s != ',') cerr << *_s++;
  cerr << " = " << _h << ",";
  _dbg(_s + 1, _t...);
}
const int64_t INF = static_cast<int64_t>(1e9) + 7;
const int64_t LINF = INF * INF;
const int MAXN = static_cast<int>(1e6) + 17;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<vector<int>> g(n);
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  int cur = 1;
  vector<int> id(n, 0), count(4, 0);
  for (int i = 0; i < n; ++i) {
    if (id[i] == 0) {
      if (cur == 4) {
        cout << "-1\n";
        return;
      }
      vector<int> ngb(n, 0);
      for (int v : g[i]) {
        ngb[v] = 1;
      }
      for (int j = 0; j < n; ++j) {
        if (ngb[j] == 0 && id[j]) {
          cout << "-1\n";
          return;
        } else if (ngb[j] == 0) {
          id[j] = cur;
          count[cur]++;
        }
      }
      ++cur;
    }
  }
  if (cur != 4) {
    cout << "-1\n";
    return;
  }
  for (int i = 0; i < n; ++i) {
    int cnt = 0;
    for (int ngb : g[i]) {
      if (id[ngb] == id[i]) {
        cout << "-1\n";
        return;
      }
      ++cnt;
    }
    if (cnt != n - count[id[i]]) {
      cout << "-1\n";
      return;
    }
  }
  for (int i = 0; i < n; ++i) {
    cout << id[i] << ' ';
  }
  cout << '\n';
}
int main() {
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios_base::sync_with_stdio(false);
  int q = 1;
  for (int i = 0; i < q; ++i) {
    solve();
  }
}
