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 fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  fast();
  long long n, m;
  cin >> n >> m;
  vector<long long> g[n + 10];
  long long color[n + 10];
  long long siz[n + 10];
  bool no1[n + 10];
  for (int i = 1; i <= n; ++i) {
    color[i] = 0;
    no1[i] = 0;
  }
  long long u, v;
  for (int i = 1; i <= m; ++i) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
    siz[u]++;
    siz[v]++;
  }
  if (siz[1] <= 1) {
    cout << -1;
    return 0;
  }
  color[1] = 1;
  v = 0;
  for (int i = 0; i < g[1].size(); ++i) no1[g[1][i]] = 1;
  for (int i = 1; i <= n; ++i)
    if (!no1[i]) color[i] = 1;
  for (int i = 1; i <= n; ++i)
    if (color[i] == 0) v = i;
  if (v == 0) {
    cout << -1;
    return 0;
  }
  if (siz[v] <= 1) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    no1[i] = 0;
  }
  for (int i = 0; i < g[v].size(); ++i) no1[g[v][i]] = 1;
  for (int i = 1; i <= n; ++i)
    if (!no1[i]) color[i] = 2;
  for (int i = 1; i <= n; ++i)
    if (!color[i]) color[i] = 3;
  long long sum[5];
  for (int i = 1; i <= 3; ++i) sum[i] = 0;
  for (int i = 1; i <= n; ++i) sum[color[i]]++;
  long long ans = sum[1] * sum[2] + sum[1] * sum[3] + sum[2] * sum[3];
  if (sum[1] == 0 || sum[2] == 0 || sum[3] == 0) {
    cout << -1;
    return 0;
  }
  if (ans != m) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    if (siz[i] <= 1) {
      cout << -1;
      return 0;
    }
    for (int j = 0; j < g[i].size(); ++j) {
      if (color[i] == color[g[i][j]]) {
        cout << -1;
        return 0;
      }
    }
  }
  for (int i = 1; i <= n; ++i) cout << color[i] << " ";
  return 0;
}
