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>
template <typename Arp1>
void ZZ(const char* name, Arp1&& arp1) {
  std::cout << name << " = " << arp1 << std::endl;
}
template <typename Arp1, typename... Args>
void ZZ(const char* names, Arp1&& arp1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cout.write(names, comma - names) << " = " << arp1;
  ZZ(comma, args...);
}
using namespace std;
long long n, m;
vector<set<long long>> adj;
vector<long long> assigned;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  adj.resize(n + 1);
  assigned.resize(n + 1, -1);
  for (long long i = 0; i < m; i++) {
    long long x, y;
    cin >> x >> y;
    adj[x].insert(y);
    adj[y].insert(x);
  }
  vector<vector<long long>> partition_members(4);
  long long current_partition = 1;
  for (long long i = 1; i <= n; i++) {
    if (assigned[i] == -1) {
      if (current_partition > 3) {
        cout << -1 << endl;
        return 0;
      }
      assigned[i] = current_partition;
      partition_members[current_partition].push_back(i);
      for (long long j = i + 1; j <= n; j++) {
        if (assigned[j] == -1 && adj[i].find(j) == adj[i].end()) {
          assigned[j] = current_partition;
          partition_members[current_partition].push_back(j);
        }
      }
      current_partition++;
    }
  }
  if (current_partition < 4 ||
      m != (partition_members[1].size() * partition_members[2].size() +
            partition_members[2].size() * partition_members[3].size() +
            partition_members[3].size() * partition_members[1].size())) {
    cout << -1 << endl;
    return 0;
  }
  for (long long p1 = 1; p1 < 3; p1++) {
    for (long long p2 = p1 + 1; p2 <= 3; p2++) {
      for (long long v1 : partition_members[p1]) {
        for (long long v2 : partition_members[p2]) {
          if (adj[v1].find(v2) == adj[v1].end()) {
            cout << -1 << endl;
            return 0;
          }
        }
      }
    }
  }
  for (long long i = 1; i <= n; i++) {
    cout << assigned[i] << " ";
  }
  cout << endl;
  return 0;
}
