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;
void printvector(std::vector<long long> v) {
  for (long long i = 0; i < v.size(); ++i) {
    cout << v[i] << " ";
  }
  cout << "" << endl;
}
void printarray(long long a[], long long n) {
  for (long long i = 0; i < n; ++i) {
    cout << a[i] << " ";
  }
  cout << "" << endl;
}
void dfs(long long s, std::set<long long> v[], vector<bool> &visited,
         long long parent, std::vector<long long> &colour) {
  visited[s] = true;
  long long fir = -1;
  for (auto adj : v[s]) {
    if (visited[adj]) continue;
    if (parent != -1) {
      if (v[adj].find(parent) != v[adj].end())
        colour[adj] = 3 - colour[s] - colour[parent];
      else
        colour[adj] = colour[parent];
    } else {
      if (fir == -1) {
        fir = adj;
        colour[adj] = 1;
      } else {
        if (v[adj].find(fir) != v[adj].end())
          colour[adj] = 3 - colour[s] - colour[fir];
        else
          colour[adj] = colour[fir];
      }
    }
    dfs(adj, v, visited, s, colour);
  }
}
void solve() {
  long long n, m;
  cin >> n >> m;
  std::set<long long> v[n];
  for (long long i = 0; i < m; i++) {
    long long a, b;
    cin >> a >> b;
    a--;
    b--;
    v[a].insert(b);
    v[b].insert(a);
  }
  std::vector<long long> colour(n, -1);
  std::vector<bool> visited(n, false);
  colour[0] = 0;
  dfs(0, v, visited, -1, colour);
  for (long long i = 0; i < n; i++) {
    bool it = visited[i];
    if (it == false) {
      cout << -1 << endl;
      return;
    }
  }
  long long no[3];
  memset(no, 0, sizeof(no));
  std::vector<vector<long long> > deg(n, std::vector<long long>(3, 0));
  for (long long i = 0; i < n; i++) {
    no[colour[i]]++;
    for (auto it : v[i]) {
      deg[i][colour[it]]++;
    }
  }
  for (long long j = 0; j < 3; j++) {
    if (no[j] == 0) {
      cout << -1 << endl;
      return;
    }
  }
  for (long long i = 0; i < n; i++) {
    std::vector<long long> temp;
    for (long long j = 0; j < 3; j++) {
      if (colour[i] == j)
        temp.push_back(0);
      else
        temp.push_back(no[j]);
    }
    if (deg[i] != temp) {
      cout << -1 << endl;
      return;
    }
  }
  for (long long i = 0; i < n; i++) {
    colour[i]++;
  }
  printvector(colour);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
