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;
const int N = 3e5 + 500;
const long long mod = 1e9 + 7;
const long long cmod = 998244353;
const long long inf = 1LL << 57;
const int M = 1e6 + 500;
const long long int ths = 1LL << 40;
vector<long long int> graph[N];
long long int col[N];
void solve() {
  long long int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    long long int u, v;
    cin >> u >> v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  for (int i = 0; i <= n; i++) col[i] = 1;
  bool ok[4] = {};
  long long int prev = 0;
  for (auto it : graph[1]) {
    col[it] = 2;
    prev = it;
  }
  for (auto it : graph[prev]) {
    if (col[it] == 2) {
      col[it] = 3;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (auto it : graph[i]) {
      if (col[i] == col[it]) {
        cout << -1;
        return;
      }
    }
  }
  long long int cnt[4] = {};
  for (int i = 1; i <= n; i++) cnt[col[i]]++;
  long long int total_edges = 0;
  total_edges += cnt[1] * cnt[2] + cnt[1] * cnt[3] + cnt[2] * cnt[3];
  for (int i = 1; i <= n; i++) {
    ok[col[i]] = 1;
  }
  if (ok[1] == 1 && ok[2] == 1 && ok[3] == 1 && total_edges == m) {
    for (int i = 1; i <= n; i++) cout << col[i] << ' ';
  } else {
    cout << -1;
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
