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;
const int mod = (int)1e9 + 7;
map<pair<int, int>, int> exist;
vector<int> g[100100];
vector<int> v[4];
int col[100100];
set<int> second;
int n, m;
vector<int> get(int v) {
  vector<int> ans;
  for (int i = 1; i <= n; ++i) {
    if (!exist.count(make_pair(v, i))) {
      if (second.find(i) != second.end()) second.erase(i);
      ans.push_back(i);
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    int x, y;
    cin >> x >> y;
    exist[make_pair(x, y)] = 1;
    exist[make_pair(y, x)] = 1;
  }
  for (int i = 1; i <= n; ++i) {
    second.insert(i);
  }
  for (int i = 0; i < 3; ++i) {
    if (second.empty()) return cout << -1 << endl, 0;
    v[i] = get(*second.begin());
    for (int j = 0; j < v[i].size(); ++j) {
      col[v[i][j]] = i + 1;
    }
  }
  if (!second.empty()) return cout << -1 << endl, 0;
  for (auto x : exist) {
    if (col[x.first.first] == col[x.first.second]) return cout << -1 << endl, 0;
  }
  if (m != v[0].size() * 1ll * v[1].size() + v[1].size() * 1ll * v[2].size() +
               v[0].size() * 1ll * v[2].size())
    return cout << -1 << endl, 0;
  for (int i = 1; i <= n; ++i) {
    cout << col[i] << " ";
  }
}
