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 long long mod = 1e9 + 7;
const long long inf = 1e9;
const long long N = 1e5 + 10;
long long mod_expo(long long a, long long b, long long c) {
  long long ans = 1;
  while (b) {
    if (b & (long long)1) ans = (ans * a) % c;
    a = (a * a) % c;
    b = b >> 1;
  }
  return ans;
}
long long inverse(long long n) { return mod_expo(n, mod - 2, mod); }
vector<int> v[N];
bool v1[N], v2[N], v3[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > edge;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
    edge.push_back(make_pair(x, y));
  }
  set<int> s;
  int a = n, b = n, c = n;
  for (int i = 1; i <= n; i++) s.insert(int(v[i].size()));
  if (int(s.size()) > 3) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i <= n; i++) v1[i] = 1;
  for (int i : v[1]) v1[i] = 0, a--;
  if (int(v[1].size()) == 0) {
    cout << -1;
    return 0;
  }
  int node = v[1][0];
  for (int i = 1; i <= n; i++) v2[i] = 1;
  for (int i : v[node]) v2[i] = 0, b--;
  node = -1;
  for (int i = 1; i <= n; i++)
    if ((!v1[i]) and (!v2[i])) {
      node = i;
      break;
    }
  if (node == -1) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i <= n; i++) v3[i] = 1;
  for (int i : v[node]) v3[i] = 0, c--;
  if ((a * b + b * c + a * c) != m) {
    cout << "-1";
    return 0;
  }
  bool flag = 1;
  vector<int> chk{0, 0, 1};
  for (int i = 1; i <= n; i++) {
    vector<int> temp;
    temp.push_back(v1[i]);
    temp.push_back(v2[i]);
    temp.push_back(v3[i]);
    sort(temp.begin(), temp.end());
    if (temp != chk) flag = 0;
  }
  if (flag == 0) {
    cout << "-1";
    return 0;
  }
  for (auto &i : edge) {
    if ((v1[i.first] and v1[i.second] == 0))
      ;
    else if ((v2[i.first] and v2[i.second] == 0))
      ;
    else if ((v3[i.first] and v3[i.second] == 0))
      ;
    else
      flag = 0;
  }
  if (flag == 0) {
    cout << "-1";
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (v1[i])
      cout << "1 ";
    else if (v2[i])
      cout << "2 ";
    else if (v3[i])
      cout << "3 ";
  }
  return 0;
}
