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>
#pragma GCC optimize("Ofast,unroll-loops")
using std::cin;
using std::cout;
using std::lower_bound;
using std::pair;
using std::sort;
using std::string;
using std::swap;
using std::unique;
using std::vector;
using int64 = long long int;
using pii = pair<int, int>;
using pii64 = pair<int64, int64>;
const int NIL = -1;
void init_IO() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
const int MAX_N = (int)1e5 + 777;
const int MAX_M = (int)3e5 + 777;
vector<int> g[MAX_N];
int uni = 0, color[MAX_N];
int n, m;
int64 cnt[17];
void solve() {
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    int v, u;
    cin >> v >> u;
    --v, --u;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  vector<pair<vector<int>, int>> ord;
  for (int v = 0; v < n; ++v) {
    sort(g[v].begin(), g[v].end());
    ord.emplace_back(g[v], v);
  }
  sort(ord.begin(), ord.end());
  for (int v = 0; v < n; ++v) {
    if (v == 0 || ord[v - 1].first != ord[v].first) {
      ++uni;
    }
    if (uni > 3) break;
    int real_vertex = ord[v].second;
    color[real_vertex] = uni;
    ++cnt[color[real_vertex]];
  }
  if (uni != 3) {
    cout << NIL << '\n';
    return;
  }
  for (int v = 0; v < n; ++v) {
    for (auto& u : g[v]) {
      if (color[v] == color[u]) {
        cout << NIL << '\n';
        return;
      }
    }
    int magic = n - cnt[color[v]];
    if (((int)(g[v]).size()) != magic) {
      cout << NIL << '\n';
      return;
    }
  }
  int64 expected_edges = 0;
  expected_edges += cnt[1] * cnt[2];
  expected_edges += cnt[2] * cnt[3];
  expected_edges += cnt[3] * cnt[1];
  int64 real_edges = m;
  if (real_edges != expected_edges) {
    cout << NIL << '\n';
    return;
  }
  if (cnt[1] == 0 || cnt[2] == 0 || cnt[3] == 0) {
    cout << NIL << '\n';
    return;
  }
  for (int v = 0; v < n; ++v) {
    cout << color[v] << ' ';
  }
  cout << '\n';
}
int main() {
  init_IO();
  solve();
  return 0;
}
