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>
#pragma GCC optimize("Ofast")
using namespace std;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<vector<int>> gr(n);
  for (int i = (int)0; i < (int)m; i++) {
    int v, u;
    cin >> v >> u, v--, u--;
    gr[v].push_back(u), gr[u].push_back(v);
  }
  vector<int> colors(n, -1);
  vector<int> cnt_colors(3);
  vector<set<int>> connected(3);
  for (int v = (int)0; v < (int)n; v++) {
    vector<int> cnt_possible_colors(3);
    for (int u : gr[v])
      if (colors[u] != -1) cnt_possible_colors[colors[u]]++;
    for (int i : {0, 1, 2}) {
      bool ok = true;
      for (int j = (int)0; j < (int)3; j++) {
        if (i == j && cnt_possible_colors[j]) ok = false;
        if (i != j) {
          if (cnt_possible_colors[j] != cnt_colors[j]) {
            ok = false;
          }
        }
      }
      if (ok) {
        colors[v] = i;
        cnt_colors[i]++;
        break;
      }
    }
    if (colors[v] == -1) return cout << -1, 0;
    for (int u : gr[v]) connected[colors[v]].insert(u);
  }
  for (int i = (int)0; i < (int)3; i++)
    if (!cnt_colors[i]) return cout << -1, 0;
  for (int i = (int)0; i < (int)n; i++) cout << colors[i] + 1 << " ";
  return 0;
}
