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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int N, M;
  cin >> N >> M;
  vector<unordered_set<int>> adj(N, unordered_set<int>());
  vector<pair<int, int>> edges;
  for (int i = 0; i < M; i++) {
    int a, b;
    cin >> a >> b;
    adj[a - 1].insert(b - 1);
    adj[b - 1].insert(a - 1);
    edges.push_back(make_pair(a - 1, b - 1));
  }
  vector<int> group(N, 1);
  for (int i : adj[0]) {
    group[i] = 2;
  }
  int v = 0;
  while (v < N && group[v] != 2) {
    v++;
  }
  if (v == N) {
    cout << "-1\n";
    return 0;
  }
  for (int i : adj[v]) {
    if (group[i] == 2) {
      group[i] = 3;
    }
  }
  for (pair<int, int> e : edges) {
    if (group[e.first] == group[e.second]) {
      cout << "-1\n";
      return 0;
    }
  }
  vector<long long> sizes(3, 0);
  for (int i = 0; i < N; i++) {
    sizes[group[i] - 1]++;
  }
  long long m = sizes[0] * sizes[1] + sizes[0] * sizes[2] + sizes[1] * sizes[2];
  if (m != M || sizes[2] == 0) {
    cout << "-1\n";
    return 0;
  }
  for (int i = 0; i < N; i++) {
    cout << group[i] << " ";
  }
  cout << '\n';
  return 0;
}
