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;
long long int mo = 1000000007;
long long int MO = 998244353;
int n, m;
vector<vector<int> > adj(100005);
int flag = 1;
int s[100005];
void foo(int i, int k) {
  vector<int> v(100005, 0);
  for (auto j : adj[i]) {
    v[j] = 1;
  }
  for (int j = 1; j <= n; j++) {
    if (!v[j]) {
      if (s[j] != 0) {
        flag = 0;
      }
      s[j] = k;
    }
  }
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    ;
    adj[y].push_back(x);
    ;
  }
  int setno = 1;
  for (int i = 1; i < n + 1; i++) {
    if (s[i] == 0) {
      foo(i, setno);
      setno++;
    }
  }
  int c1 = 0, c2 = 0, c3 = 0;
  for (int i = 1; i < n + 1; i++) {
    if (s[i] == 1) c1++;
    if (s[i] == 2) c2++;
    if (s[i] == 3) c3++;
  }
  if (!c1 || !c2 || !c3 || c1 + c2 + c3 != n ||
      c1 * c2 + c2 * c3 + c3 * c1 != m) {
    flag = 0;
  }
  for (int i = 1; i <= n; i++) {
    for (auto j : adj[i]) {
      if (s[j] == s[i]) {
        flag = 0;
        break;
      }
    }
  }
  if (!flag)
    cout << -1;
  else {
    for (int i = 1; i < n + 1; i++) {
      cout << s[i] << " ";
    }
  }
  cout << endl;
  return 0;
}
