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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
const int mxN = 3e5 + 5;
template <class T>
T gcd(T a, T b) {
  return ((b == 0) ? a : gcd(b, a % b));
}
vector<int> g[mxN];
int col[mxN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < (int)(m); i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  set<int> st(g[0].begin(), g[0].end());
  int f = 0;
  int num1 = 0, num2 = 0, num3 = 0;
  for (int i = 0; i < (int)(n); i++)
    if (st.find(i) == st.end()) {
      col[i] = 1;
      f |= 1;
      num1++;
    }
  int gd = -1;
  for (int i = 0; i < (int)(n); i++)
    if (!col[i]) gd = i;
  set<int> st1(g[gd].begin(), g[gd].end());
  for (int i = 0; i < (int)(n); i++)
    if (st1.find(i) == st1.end()) {
      if (col[i]) {
        cout << -1 << endl;
        return 0;
      }
      col[i] = 2;
      num2++;
      f |= 2;
    }
  for (int i = 0; i < (int)(n); i++)
    if (!col[i]) col[i] = 3, f |= 4, num3++;
  if (f < 7) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0; i < (int)(n); i++) {
    int nm1 = 0, nm2 = 0, nm3 = 0;
    for (int j : g[i]) {
      if (col[j] == 1) nm1++;
      if (col[j] == 2) nm2++;
      if (col[j] == 3) nm3++;
    }
    if (col[i] == 1) {
      if (nm1 || nm2 != num2 || nm3 != num3) {
        cout << -1 << endl;
        return 0;
      }
    }
    if (col[i] == 2) {
      if (nm1 != num1 || nm2 || nm3 != num3) {
        cout << -1 << endl;
        return 0;
      }
    }
    if (col[i] == 3) {
      if (nm1 != num1 || nm2 != num2 || nm3) {
        cout << -1 << endl;
        return 0;
      }
    }
  }
  for (int i = 0; i < (int)(n); i++) cout << col[i] << " ";
  cout << endl;
  return 0;
}
