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 char el = '\n';
int deg[100001];
int assignment[100001];
bool seen[100001];
int other(int x, int y) {
  vector<bool> v(4, true);
  v[x] = false;
  v[y] = false;
  for (int i = 1; i <= 3; ++i) {
    if (v[i]) return i;
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  vector<set<int>> nbr(n + 1);
  fill(deg + 1, deg + 1 + n, 0);
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    ++deg[u], ++deg[v];
    nbr[u].emplace(v);
    nbr[v].emplace(u);
  }
  vector<int> numwithassignment(4, 0);
  int assigned = 0;
  assignment[1] = 1;
  ++numwithassignment[1];
  seen[1] = true;
  ++assigned;
  if (nbr[1].empty()) {
    cout << -1 << el;
    return 0;
  }
  stack<pair<int, int>> s;
  int temp = *nbr[1].begin();
  assignment[temp] = 2;
  ++numwithassignment[2];
  seen[temp] = true;
  ++assigned;
  s.push({temp, 1});
  for (int j : nbr[1]) {
    if (j == temp) continue;
    if (nbr[temp].count(j)) {
      assignment[j] = 3;
    } else {
      assignment[j] = 2;
    }
    ++numwithassignment[assignment[j]];
    seen[j] = true;
    ++assigned;
    s.push({j, 1});
  }
  while (!s.empty()) {
    auto pr = s.top();
    s.pop();
    int i = pr.first, j = pr.second;
    for (int k : nbr[i]) {
      if (seen[k]) {
        if (assignment[k] == assignment[i]) {
          cout << -1 << el;
          return 0;
        } else if (nbr[j].count(k)) {
          if (assignment[k] == assignment[j]) {
            cout << -1 << el;
            return 0;
          }
        } else {
          if (assignment[k] != assignment[j]) {
            cout << -1 << el;
            return 0;
          }
        }
      } else {
        if (nbr[j].count(k)) {
          assignment[k] = other(assignment[j], assignment[i]);
        } else {
          assignment[k] = assignment[j];
        }
        ++numwithassignment[assignment[k]];
        s.push({k, i});
        seen[k] = true;
        ++assigned;
      }
    }
  }
  if (assigned < n) {
    cout << -1 << el;
    return 0;
  }
  for (int i = 1; i <= 3; ++i) {
    if (numwithassignment[i] == 0) {
      cout << -1 << el;
      return 0;
    }
  }
  vector<int> complement(4, 0);
  complement[1] = numwithassignment[2] + numwithassignment[3];
  complement[2] = numwithassignment[1] + numwithassignment[3];
  complement[3] = numwithassignment[1] + numwithassignment[2];
  for (int u = 1; u <= n; ++u) {
    int j = assignment[u];
    if (deg[u] != complement[j]) {
      cout << -1 << el;
      return 0;
    }
  }
  for (int u = 1; u <= n; ++u) {
    cout << assignment[u] << " ";
  }
  cout << el;
}
