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 print1() {
  cout << -1 << endl;
  return 0;
}
int main() {
  int n, m;
  cin >> n >> m;
  vector<vector<int> > e(n);
  int a, b;
  for (int i = 0; i < m; ++i) {
    cin >> a >> b;
    a--;
    b--;
    e[a].push_back(b);
    e[b].push_back(a);
  }
  vector<int> c(n, 0);
  vector<int> cnt(3, 0);
  cnt[0] = n;
  for (auto &v : e[0]) {
    c[v] = 2;
    cnt[0]--;
    cnt[2]++;
  }
  for (a = 0; a < n && c[a] == 0; ++a)
    ;
  if (a == n) return print1();
  for (auto &v : e[a]) {
    if (c[v] == 2) {
      c[v] = 1;
      cnt[1]++;
      cnt[2]--;
    }
  }
  for (int j = 0; j < 3; ++j)
    if (cnt[j] == 0) return print1();
  vector<int> loc(3, 0);
  for (int i = 0; i < n; ++i) {
    loc[0] = 0;
    loc[1] = 0;
    loc[2] = 0;
    for (auto &v : e[i]) loc[c[v]]++;
    if (loc[c[i]] != 0) return print1();
    for (int j = 0; j < 3; ++j)
      if (c[i] != j && loc[j] != cnt[j]) return print1();
  }
  for (int i = 0; i < n; ++i) cout << c[i] + 1 << " ";
  cout << endl;
  return 0;
}
