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 n, m, md[100003], vis[100003], ncnt[10];
vector<int> graph[100003];
pair<int, int> edge[300003];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    graph[u].push_back(v);
    graph[v].push_back(u);
    edge[i] = {u, v};
  }
  int c = 1;
  for (int i = 1; i <= n; i++) {
    if (!md[i]) {
      if (c > 3) {
        cout << -1 << endl;
        return 0;
      }
      for (auto &nxt : graph[i]) {
        vis[nxt] = c;
      }
      for (int j = 1; j <= n; j++) {
        if (vis[j] != c) {
          if (md[j]) {
            cout << -1 << endl;
            return 0;
          }
          md[j] = c;
        }
      }
      c++;
    }
  }
  if (c <= 3) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    ncnt[md[i]]++;
  }
  map<pair<int, int>, int> mp;
  for (int i = 0; i < m; i++) {
    int u, v;
    tie(u, v) = edge[i];
    if (md[u] == md[v]) {
      cout << -1 << endl;
      return 0;
    }
    int c1 = md[u], c2 = md[v];
    if (c1 > c2) swap(c1, c2);
    mp[{c1, c2}]++;
  }
  for (int i = 1; i < 3; i++) {
    for (int j = i + 1; j <= 3; j++) {
      if (1LL * mp[{i, j}] != 1LL * ncnt[i] * ncnt[j]) {
        cout << -1 << endl;
        return 0;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (i > 1) cout << ' ';
    cout << md[i];
  }
  cout << endl;
  return 0;
}
