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 target("avx2")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long power(long long x, long long y, long long m) {
  long long temp;
  if (y == 0) return 1;
  temp = (power(x, y / 2, m)) % m;
  if (y % 2 == 0)
    return ((temp % m) * temp);
  else
    return ((x * temp % m) * temp % m) % m;
}
long long inv(long long x) {
  return (power(x, 998244353 - 2, 998244353)) % 998244353;
}
bool comp(long long a, long long b) { return a > b; }
long long parity[100001], q = 0, freq[3];
vector<vector<long long>> edges(300001);
vector<set<long long>> edgeslook(300001);
void QFS(long long v) {
  parity[v] = 0;
  for (long long i = 0; i < edges[v].size(); i++) {
    if (!i) {
      parity[edges[v][i]] = 1;
      q = edges[v][i];
    } else {
      if (edgeslook[q].find(edges[v][i]) != edgeslook[q].end())
        parity[edges[v][i]] = 2;
      else
        parity[edges[v][i]] = 1;
    }
  }
  for (long long i = 0; i < edges[q].size(); i++) {
    if (edges[q][i] == v) continue;
    if (edgeslook[v].find(edges[q][i]) == edgeslook[v].end())
      parity[edges[q][i]] = 0;
  }
}
int32_t main() {
  iostream::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, a, b;
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    cin >> a >> b;
    edges[a - 1].push_back(b - 1), edges[b - 1].push_back(a - 1);
    edgeslook[a - 1].insert(b - 1), edgeslook[b - 1].insert(a - 1);
  }
  for (long long i = 0; i < n; i++) parity[i] = -1;
  QFS(0);
  vector<long long> answer;
  for (long long i = 0; i < n; i++) {
    if (parity[i] == -1) return cout << -1, 0;
    freq[parity[i]]++;
  }
  if (!(freq[0] && freq[1] && freq[2])) return cout << -1, 0;
  for (long long i = 0; i < n; i++) {
    for (long long l = 0; l < edges[i].size(); l++) {
      if (parity[edges[i][l]] == parity[i]) return cout << -1, 0;
    }
    if (edges[i].size() != (freq[0] + freq[1] + freq[2] - freq[parity[i]]))
      return cout << -1, 0;
  }
  for (long long i = 0; i < n; i++) cout << parity[i] + 1 << ' ';
  return 0;
}
