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 gcd(int a, int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
int main() {
  ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  vector<vector<int> > edges(n + 10, vector<int>());
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    edges[x].push_back(y);
    edges[y].push_back(x);
  }
  map<int, int> not_first;
  map<int, int> not_second;
  map<int, int> not_third;
  set<int> first;
  set<int> second;
  set<int> third;
  for (int i = 1; i < n + 1; i++) {
    if (not_first.find(i) == not_first.end()) {
      first.insert(i);
      for (int j = 0; j < edges[i].size(); j++) {
        not_first[edges[i][j]]++;
      }
    } else if (not_second.find(i) == not_second.end()) {
      second.insert(i);
      for (int j = 0; j < edges[i].size(); j++) {
        not_second[edges[i][j]]++;
      }
    } else if (not_third.find(i) == not_third.end()) {
      third.insert(i);
      for (int j = 0; j < edges[i].size(); j++) {
        not_third[edges[i][j]]++;
      }
    } else {
      cout << -1;
      return 0;
    }
  }
  if (second.size() == 0 or third.size() == 0) {
    cout << -1;
  } else {
    for (auto i = not_first.begin(); i != not_first.end(); i++) {
      if (i->second != first.size()) {
        cout << -1;
        return 0;
      }
    }
    for (auto i = not_second.begin(); i != not_second.end(); i++) {
      if (i->second != second.size()) {
        cout << -1;
        return 0;
      }
    }
    for (auto i = not_third.begin(); i != not_third.end(); i++) {
      if (i->second != third.size()) {
        cout << -1;
        return 0;
      }
    }
    vector<int> cols(n + 1, 0);
    for (auto i = first.begin(); i != first.end(); i++) {
      cols[*i] = 1;
    }
    for (auto i = second.begin(); i != second.end(); i++) {
      cols[*i] = 2;
    }
    for (auto i = third.begin(); i != third.end(); i++) {
      cols[*i] = 3;
    }
    for (int i = 1; i < n + 1; i++) {
      cout << cols[i] << " ";
    }
  }
  return 0;
}
