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 <class T>
inline T updmax(T& a, T b) {
  return a = max(a, b);
}
template <class T>
inline T updmin(T& a, T b) {
  return a = min(a, b);
}
using edge = int;
using node = vector<edge>;
using graph = vector<node>;
void addedge(graph& g, int from, int to, bool bidirectional = false) {
  g[from].emplace_back(edge(to));
  if (bidirectional) g[to].emplace_back(edge(from));
}
class Solution {
 public:
  void solve() {
    int n, m;
    cin >> n >> m;
    vector<pair<int, int>> es;
    graph g(n);
    for (int i = 0; i < m; i++) {
      int a, b;
      cin >> a >> b;
      a--, b--;
      addedge(g, a, b, true);
    }
    vector<int> id(n, -1);
    vector<long long> cnt;
    int iid = 0;
    for (int i = 0, j = 0; i < n; i++) {
      if (id[i] != -1) continue;
      int tmp_cnt = 0;
      sort(g[i].begin(), g[i].end());
      g[i].emplace_back(n);
      for (int j = 0, x = 0; j < g[i].size(); j++, x++) {
        while (x != g[i][j]) {
          if (id[x] != -1) {
            cout << -1 << endl;
            return;
          }
          id[x++] = iid;
          tmp_cnt++;
        }
      }
      g[i].pop_back();
      iid++;
      cnt.emplace_back(tmp_cnt);
    }
    if (iid != 3) {
      cout << -1 << endl;
      return;
    }
    for (int i = 0; i < n; i++) {
      if (g[i].size() != n - cnt[id[i]]) {
        cout << -1 << endl;
        return;
      }
      for (int j = 0; j < g[i].size(); j++) {
        if (id[i] == id[g[i][j]]) {
          cout << -1 << endl;
          return;
        }
      }
    }
    for (int i = 0; i < n; i++) cout << id[i] + 1 << " ";
    cout << "\n";
    return;
  };
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  Solution solution;
  solution.solve();
  return 0;
}
