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 A>
ostream &operator<<(ostream &cout, vector<A> const &v);
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << "(" << p.f << ", " << p.s << ")";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
template <typename A, typename B>
istream &operator>>(istream &cin, pair<A, B> &p) {
  cin >> p.first;
  return cin >> p.second;
}
vector<set<long long>> adj;
void solve() {
  long long v, e;
  cin >> v >> e;
  if (e <= 0) {
    cout << -1 << '\n';
    return;
  }
  long long n = v;
  adj = vector<set<long long>>(v + 5);
  set<long long> fs1, fs2, fs3;
  set<long long> s1, temps2;
  for (long long i = 0, u, v; i < e; i++) {
    cin >> u >> v;
    adj[u].insert(v);
    adj[v].insert(u);
    s1.insert(u);
    s1.insert(v);
  }
  for (long long i : adj[*s1.begin()]) {
    s1.erase(i);
    temps2.insert(i);
  }
  fs1 = s1;
  for (long long i : temps2) {
    for (long long x : adj[i]) {
      if (fs1.find(x) == fs1.end()) {
        fs3.insert(x);
        temps2.erase(x);
      }
    }
  }
  fs2 = temps2;
  bool ok = true;
  for (int i : fs1) {
    for (int j : fs2) {
      if (adj[i].find(j) == adj[i].end()) {
        ok = false;
        break;
      }
    }
  }
  for (int i : fs1) {
    for (int j : fs3) {
      if (adj[i].find(j) == adj[i].end()) {
        ok = false;
        break;
      }
    }
  }
  for (int i : fs2) {
    for (int j : fs3) {
      if (adj[i].find(j) == adj[i].end()) {
        ok = false;
        break;
      }
    }
  }
  for (long long i = 1; i <= n && ok; i++) {
    if (fs1.find(i) != fs1.end()) {
      for (long long x : adj[i]) {
        long long counter =
            (fs2.find(x) != fs2.end()) + (fs3.find(x) != fs3.end());
        if (counter ^ 1) {
          ok = false;
          break;
        }
      }
    } else if (fs2.find(i) != fs2.end()) {
      for (long long x : adj[i]) {
        long long counter =
            (fs1.find(x) != fs1.end()) + (fs3.find(x) != fs3.end());
        if (counter ^ 1) {
          ok = false;
          break;
        }
      }
    } else if (fs3.find(i) != fs3.end()) {
      for (long long x : adj[i]) {
        long long counter =
            (fs1.find(x) != fs1.end()) + (fs2.find(x) != fs2.end());
        if (counter ^ 1) {
          ok = false;
          break;
        }
      }
    } else {
      ok = false;
    }
  }
  if (fs1.size() == 0 || fs2.size() == 0 || fs3.size() == 0 ||
      (fs1.size() + fs2.size() + fs3.size()) != v) {
    ok = false;
  }
  if (ok) {
    for (long long i = 1; i <= n; i++) {
      if (fs1.find(i) != fs1.end()) {
        cout << 1 << " ";
      }
      if (fs2.find(i) != fs2.end()) {
        cout << 2 << " ";
      }
      if (fs3.find(i) != fs3.end()) {
        cout << 3 << " ";
      }
    }
    cout << '\n';
  } else {
    cout << -1 << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int testcase = 1;
  for (int i = 0; i < testcase; i++) {
    solve();
  }
  return 0;
}
