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;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<set<int>> edge(n + 1);
  for (int i = 1; i <= m; ++i) {
    int u, v;
    cin >> u >> v;
    edge[u].insert(v);
    edge[v].insert(u);
  }
  vector<int> ans(n + 1);
  for (int type = 1; type <= 3; ++type) {
    int i;
    for (i = 1; i <= n; ++i)
      if (!ans[i]) break;
    if (i == n + 1) return cout << "-1", 0;
    ans[i] = type;
    for (int id = 1; id <= n; ++id) {
      if (!ans[id] && id != i && edge[i].find(id) == edge[i].end())
        ans[id] = type;
    }
  }
  vector<vector<int>> vt(3);
  for (int i = 1; i <= n; ++i) {
    if (!ans[i]) return cout << "-1", 0;
    vt[ans[i] - 1].emplace_back(i);
  }
  int counter = 0;
  auto check = [&](const vector<int>& a, const vector<int>& b) {
    for (auto& x : a)
      for (auto& y : b)
        if (edge[x].find(y) == edge[x].end()) {
          cout << "-1";
          exit(0);
        } else
          counter++;
  };
  check(vt[0], vt[1]);
  check(vt[2], vt[1]);
  check(vt[0], vt[2]);
  if (counter != m)
    cout << "-1";
  else
    for (int i = 1; i <= n; ++i) cout << ans[i] << ' ';
}
