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;
vector<long long int> lst[100005];
long long int deg[100005], c = 0, comp[100005];
long long int f(long long int n) {
  long long int i, j, k;
  long long int vis[100005] = {0};
  for (i = 1; i <= n; i++) {
    if (vis[i] == 0) {
      comp[i] = c;
      long long int same[n + 1];
      for (j = 1; j <= n; j++) same[j] = 1;
      same[i] = 0;
      unordered_set<long long int> s;
      for (j = 0; j < lst[i].size(); j++) {
        long long int to = lst[i][j];
        s.insert(to);
        same[to] = 0;
      }
      for (j = 1; j <= n; j++) {
        if (same[j]) {
          if (deg[j] != deg[i]) return -1;
          for (k = 0; k < lst[j].size(); k++) {
            if (s.find(lst[j][k]) == s.end()) {
              return -1;
            }
          }
          comp[j] = c;
          vis[j] = 1;
        }
      }
      c++;
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int i, j, n, m, u, v;
  cin >> n >> m;
  for (i = 0; i < m; i++) {
    cin >> u >> v;
    lst[u].push_back(v);
    lst[v].push_back(u);
    deg[u]++;
    deg[v]++;
  }
  long long int x = f(n);
  if (x == -1 || c != 3)
    cout << -1 << "\n";
  else {
    for (i = 1; i <= n; i++) cout << comp[i] + 1 << " ";
    cout << "\n";
  }
  return 0;
}
