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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long mod = 1e9 + 7;
void add(long long& a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
}
long long mul(long long a, long long b) { return (a * b) % mod; }
long long mulmod(long long a, long long b, long long M) {
  if (M <= mod) return mul(a, b);
  long long res = 0;
  while (a > 0) {
    if (a & 1ll) {
      res += b;
      if (res >= M) res -= M;
    }
    a >>= 1;
    b <<= 1;
    if (b >= M) b -= M;
  }
  return res;
}
long long ffmod(long long b, long long exp, long long md = mod) {
  long long res = 1;
  while (exp) {
    if (exp & 1ll) res = (res * b) % md;
    b = (b * b) % md;
    exp /= 2ll;
  }
  return res;
}
void mxupd(pair<int, int>& a, pair<int, int> b) {
  if (a.first > b.first)
    a.second = max(a.second, b.first);
  else {
    a.second = max(a.first, b.second);
    a.first = b.first;
  }
}
const int N = 200010;
const int M = 200010;
const int LN = 20;
vector<int> adj[N];
int mark[N];
int vis[N];
vector<int> v;
int main() {
  int n, m, i, j, k, x, q;
  int t = 1;
  int u, ve;
  cin >> n >> m;
  for (i = 1; i <= m; i++) {
    cin >> u >> ve;
    adj[u].push_back(ve);
    adj[ve].push_back(u);
  }
  int fl = 0;
  for (i = 1; i <= n; i++) sort(adj[i].begin(), adj[i].end());
  for (int ve : adj[1]) mark[ve] = 1;
  for (i = 1; i <= n; i++) {
    mark[i] ^= 1;
    if (mark[i]) v.push_back(i);
  }
  for (i = 1; i <= v.size() - 1; i++) {
    if (adj[v[i]] != adj[v[i - 1]]) fl = 1;
  }
  if (fl) {
    cout << "-1\n";
    return 0;
  }
  int ind = -1;
  for (i = 1; i <= n; i++)
    if (!mark[i]) {
      ind = i;
      break;
    }
  if (ind == -1) {
    cout << "-1\n";
    return 0;
  }
  v.clear();
  for (int ve : adj[ind]) {
    vis[ve] = 1;
  }
  for (i = 1; i <= n; i++) {
    if (!vis[i]) v.push_back(i);
  }
  for (i = 1; i <= v.size() - 1; i++) {
    if (adj[v[i]] != adj[v[i - 1]]) {
      fl = 1;
      break;
    }
  }
  if (fl) {
    cout << "-1\n";
    return 0;
  }
  for (int ve : v) mark[ve] = 2;
  v.clear();
  for (i = 1; i <= n; i++) {
    if (!mark[i]) {
      v.push_back(i);
      mark[i] = 3;
    }
  }
  if (v.empty()) {
    cout << "-1\n";
    return 0;
  }
  for (i = 1; i <= v.size() - 1; i++) {
    if (adj[v[i]] != adj[v[i - 1]]) {
      fl = 1;
      break;
    }
  }
  if (fl) {
    cout << "-1\n";
    return 0;
  }
  for (i = 1; i <= n; i++) cout << mark[i] << " ";
  cout << "\n";
  return 0;
}
