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 int INF = 0x3f3f3f3f;
const long long int LINF = 0x3f3f3f3f3f3f3f;
set<int> adj[100009];
int st[100009];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) st[i] = 7;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].insert(v);
    adj[v].insert(u);
  }
  vector<int> gid(n, -1);
  vector<int> grp[4];
  for (int g = 1; g <= 3; g++) {
    int fst;
    for (int i = 0; i < n; i++) {
      if (gid[i] == -1) {
        fst = i;
        break;
      }
    }
    grp[g].push_back(fst);
    gid[fst] = g;
    for (int i = 0; i < n; i++) {
      if (gid[i] != -1 || i == fst) continue;
      if (adj[fst].find(i) == adj[fst].end()) {
        gid[i] = g;
        grp[g].push_back(i);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (gid[i] == -1) {
      cout << -1 << "\n";
      return 0;
    }
  }
  int cnt = 0;
  for (int g1 = 1; g1 <= 3; g1++)
    for (int g2 = g1 + 1; g2 <= 3; g2++)
      for (int v1 : grp[g1])
        for (int v2 : grp[g2]) {
          if (adj[v1].find(v2) == adj[v1].end()) {
            cout << -1 << "\n";
            return 0;
          }
          cnt++;
        }
  if (cnt != m) {
    cout << -1 << "\n";
    return 0;
  }
  for (int i = 0; i < gid.size(); i++)
    cout << gid[i] << " \n"[i == gid.size() - 1];
}
