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;
vector<set<int> > g(200000);
vector<bool> us(200000);
vector<vector<int> > ans(3), g1(200000);
set<int> st, st1, st2, st0;
vector<int> s(200000);
int main() {
  ios::sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    g[y].insert(x);
    g[x].insert(y);
    g1[y].push_back(x);
    g1[x].push_back(y);
  }
  for (int i = 0; i < n; i++)
    if (g[0].find(i) == g[0].end()) {
      us[i] = true;
      ans[0].push_back(i);
      st.insert(i);
      st0.insert(i);
    }
  int z = 0;
  for (int i = 0; i < n; i++)
    if (!us[i]) z = i;
  for (int i = 0; i < n; i++)
    if (g[z].find(i) == g[z].end()) {
      us[i] = true;
      ans[1].push_back(i);
      if (st.find(i) != st.end()) {
        cout << -1 << endl;
        return 0;
      }
      st.insert(i);
      st1.insert(i);
    }
  for (int i = 0; i < n; i++)
    if (!us[i]) z = i;
  ;
  for (int i = 0; i < n; i++)
    if (g[z].find(i) == g[z].end()) {
      us[i] = true;
      ans[2].push_back(i);
      if (st.find(i) != st.end()) {
        cout << -1 << endl;
        return 0;
      }
      st.insert(i);
      st2.insert(i);
    }
  bool o = true;
  for (int i = 0; i < ans[0].size(); i++)
    for (int j = 0; j < g[ans[0][i]].size(); j++)
      if (st0.find(g1[ans[0][i]][j]) != st0.end()) o = false;
  for (int i = 0; i < ans[1].size(); i++)
    for (int j = 0; j < g[ans[1][i]].size(); j++)
      if (st1.find(g1[ans[1][i]][j]) != st1.end()) o = false;
  for (int i = 0; i < ans[2].size(); i++)
    for (int j = 0; j < g[ans[2][i]].size(); j++)
      if (st2.find(g1[ans[2][i]][j]) != st2.end()) o = false;
  for (int i = 0; i < n; i++)
    if (!us[i]) o = us[i];
  for (int j = 0; j < 3; j++)
    for (int i = 0; i < ans[j].size(); i++)
      if (g[ans[j][i]].size() != n - ans[j].size()) o = false;
  if (ans[0].size() == 0 || ans[1].size() == 0 || ans[2].size() == 0) o = false;
  if (o) {
    vector<int> x(n);
    for (int j = 0; j < 3; j++)
      for (int i = 0; i < ans[j].size(); i++) x[ans[j][i]] = j + 1;
    for (int i = 0; i < n; i++) cout << x[i] << ' ';
    return 0;
  }
  cout << -1;
}
