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;
set<int> get_group(int N, vector<vector<int> >& adj, int u) {
  set<int> ret;
  ret.insert(u);
  set<int> S(adj[u].begin(), adj[u].end());
  for (int v = 0, _n = (N); v < _n; ++v) {
    if (S.count(v)) continue;
    ret.insert(v);
  }
  return ret;
}
bool intersects(set<int>& g1, set<int>& g2) {
  vector<int> V;
  set_intersection(g1.begin(), g1.end(), g2.begin(), g2.end(),
                   back_inserter(V));
  return !V.empty();
}
bool has_edges(int N, vector<vector<int> >& adj, set<int>& g) {
  for (int u : g) {
    for (int v : adj[u]) {
      if (g.count(v)) return true;
    }
  }
  return false;
}
bool is_complete(int N, vector<vector<int> >& adj, set<int>& g1, set<int>& g2) {
  for (int u : g1) {
    int cnt = 0;
    for (int v : adj[u]) {
      if (g2.count(v)) ++cnt;
    }
    if (cnt != int((g2).size())) return false;
  }
  for (int u : g2) {
    int cnt = 0;
    for (int v : adj[u]) {
      if (g1.count(v)) ++cnt;
    }
    if (cnt != int((g1).size())) return false;
  }
  return true;
}
void solve(int N, vector<vector<int> >& adj) {
  int uu = -1, vv = -1;
  for (int u = 0, _n = (N); u < _n; ++u) {
    if (adj[u].empty()) continue;
    uu = u;
    vv = adj[u].front();
    break;
  }
  if (uu == -1) {
    cout << "-1\n";
    return;
  }
  set<int> group1 = get_group(N, adj, uu);
  set<int> group2 = get_group(N, adj, vv);
  if (has_edges(N, adj, group1) || has_edges(N, adj, group2)) {
    cout << "-1\n";
    return;
  }
  if (intersects(group1, group2)) {
    cout << "-1\n";
    return;
  }
  set<int> group3;
  for (int u = 0, _n = (N); u < _n; ++u) {
    if (group1.count(u) || group2.count(u)) continue;
    group3.insert(u);
  }
  if (group3.empty() || has_edges(N, adj, group3)) {
    cout << "-1\n";
    return;
  }
  if (!is_complete(N, adj, group1, group2) ||
      !is_complete(N, adj, group1, group3) ||
      !is_complete(N, adj, group2, group3)) {
    cout << "-1\n";
    return;
  }
  for (int u = 0, _n = (N); u < _n; ++u) {
    if (u > 0) cout << ' ';
    if (group1.count(u))
      cout << 1;
    else if (group2.count(u))
      cout << 2;
    else
      cout << 3;
  }
  cout << '\n';
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int N, M;
  cin >> N >> M;
  vector<vector<int> > adj(N);
  for (int j = 0, _n = (M); j < _n; ++j) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  solve(N, adj);
  return 0;
}
