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;
using ll = long long;
using ld = long double;
void no() {
  cout << -1 << endl;
  exit(0);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  if (m == 0) no();
  vector<set<int>> g(n);
  vector<int> s;
  for (int i = 0; i < m; i++) {
    int v, u;
    cin >> v >> u;
    v--, u--;
    g[v].insert(u);
    g[u].insert(v);
    if (i == 0) {
      s.push_back(v);
      s.push_back(u);
    }
  }
  for (auto u : g[s[0]]) {
    if (u == s[1]) continue;
    if (g[s[1]].find(u) != g[s[1]].end()) {
      s.push_back(u);
      break;
    }
  }
  if (int((s).size()) == 2) no();
  vector<int> color(n);
  for (int v = 0; v < n; v++) {
    int c = 0;
    for (int i = 1; i < 3; i++) {
      if (g[v].find(s[i]) == g[v].end()) c = i;
    }
    color[v] = c;
  }
  for (int v = 0; v < n; v++) {
    for (auto u : g[v]) {
      if (color[v] == color[u]) no();
    }
  }
  vector<vector<int>> groups(3);
  for (int v = 0; v < n; v++) groups[color[v]].push_back(v);
  for (int i = 0; i < 3; i++) {
    for (int j = i + 1; j < 3; j++) {
      for (auto v : groups[i]) {
        for (auto u : groups[j]) {
          if (g[v].find(u) == g[v].end()) no();
        }
      }
    }
  }
  for (auto c : color) cout << (c + 1) << ' ';
  cout << endl;
}
