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 <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << '=' << h << '\n';
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',';
  _dbg(sdbg + 1, a...);
}
const int N = 1e5 + 7;
int n, m, deg[N], v[N];
vector<int> s1, s2, s3;
vector<pair<int, int> > ed;
bool e1[N], ep[N];
void NO() {
  cout << -1 << '\n';
  exit(0);
}
void printAns() {
  if (s1.size() == 0 || s2.size() == 0 || s3.size() == 0) NO();
  for (int i = 1; i <= n; i++) {
    assert(v[i] != 0);
    cout << v[i] << " ";
  }
  cout << '\n';
  exit(0);
}
void check(vector<int> &s, int id) {
  int L = n - s.size();
  bool valid = true;
  for (auto e : ed) {
    if (v[e.first] == id && v[e.second] == id) valid = false;
  }
  for (auto v : s)
    if (deg[v] != L) valid = false;
  if (!valid) NO();
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    if (a > b) swap(a, b);
    ed.push_back(make_pair(a, b));
    if (a == 1) e1[b] = true;
    deg[a]++, deg[b]++;
  }
  s1.push_back(1);
  v[1] = 1;
  for (int i = 2; i <= n; i++)
    if (!e1[i]) s1.push_back(i), v[i] = 1;
  check(s1, 1);
  int p = -1;
  for (int i = 1; i <= n; i++)
    if (v[i] != 1) {
      p = i;
      break;
    }
  if (p == -1) printAns();
  s2.push_back(p);
  v[p] = 2;
  for (auto &e : ed) {
    if (e.second == p) swap(e.first, e.second);
    if (e.first == p) ep[e.second] = true;
  }
  for (int i = 1; i <= n; i++)
    if (!ep[i] && i != p) {
      assert(v[i] == 0);
      s2.push_back(i);
      v[i] = 2;
    }
  check(s2, 2);
  for (int i = 1; i <= n; i++)
    if (v[i] == 0) s3.push_back(i), v[i] = 3;
  check(s3, 3);
  for (auto e : ed) {
    assert(v[e.first] != v[e.second]);
  }
  printAns();
  return 0;
}
