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 F, typename S>
ostream& operator<<(ostream& os, const pair<F, S>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& os, vector<T> const& x) {
  os << "{ ";
  for (auto& y : x) os << y << " ";
  return os << "}";
}
template <class T>
ostream& operator<<(ostream& os, set<T> const& x) {
  os << "{ ";
  for (auto& y : x) os << y << " ";
  return os << "}";
}
template <typename T>
ostream& operator<<(ostream& os, const multiset<T>& v) {
  os << "[";
  typename multiset<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename F, typename S>
ostream& operator<<(ostream& os, const map<F, S>& v) {
  os << "[";
  typename map<F, S>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
const long long int MOD = 1e9 + 7;
const int N = 3e5 + 100;
long long int hh(long long int a, long long int b) {
  return ((a) < (b) ? (a) : (b)) * ((long long int)1e7) +
         ((a) > (b) ? (a) : (b));
}
int cnt[N];
unordered_set<long long int> se;
bool check1(vector<int>& v, int sz) {
  for (auto i : v)
    if (cnt[i] != sz) return false;
  return true;
}
bool check2(vector<int>& v) {
  int n = v.size();
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (se.find(hh(v[i], v[j])) != se.end()) return false;
    }
  }
  return true;
}
void _main_main() {
  long long int n, m, x, y;
  cin >> n >> m;
  while (m--) {
    cin >> x >> y;
    se.insert(hh(x, y));
    cnt[x]++;
    cnt[y]++;
  }
  vector<int> v1, v2, v3, v4;
  v1.push_back(1);
  for (int i = 2; i <= n; i++) {
    if (se.find(hh(1ll, i)) != se.end())
      v2.push_back(i);
    else
      v1.push_back(i);
  }
  if (v2.size() < 2) {
    cout << -1 << "\n";
    return;
  }
  for (auto i : v2) {
    if (se.find(hh(i, v2[0])) != se.end()) {
      v4.emplace_back(i);
    } else
      v3.emplace_back(i);
  }
  if (v3.empty()) {
    cout << -1 << "\n";
    return;
  }
  if (v4.empty()) {
    cout << -1 << "\n";
    return;
  }
  if (!check1(v1, v3.size() + v4.size())) {
    cout << -1 << "\n";
    return;
  }
  if (!check1(v3, v1.size() + v4.size())) {
    cout << -1 << "\n";
    return;
  }
  if (!check1(v4, v1.size() + v3.size())) {
    cout << -1 << "\n";
    return;
  }
  vector<pair<int, int> > pp;
  pp.push_back({v1.size(), 1});
  pp.push_back({v3.size(), 3});
  pp.push_back({v4.size(), 4});
  sort(pp.begin(), pp.end());
  for (int i = 0; i < 2; i++) {
    if (pp[i].second == 1) {
      if (!check2(v1)) {
        cout << -1 << "\n";
        return;
      }
    }
    if (pp[i].second == 3) {
      if (!check2(v3)) {
        cout << -1 << "\n";
        return;
      }
    }
    if (pp[i].second == 4) {
      if (!check2(v4)) {
        cout << -1 << "\n";
        return;
      }
    }
  }
  long long int ans[n + 5];
  for (auto i : v1) ans[i] = 1;
  for (auto i : v3) ans[i] = 2;
  for (auto i : v4) ans[i] = 3;
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int testCase = 1;
  for (int i = 0; i < testCase; i++) {
    _main_main();
  }
}
