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;
set<int> a[100100];
void check(vector<int> &group0, vector<int> &group1) {
  for (int x : group0)
    for (int y : group1)
      if (!a[x].count(y)) {
        cout << -1 << endl;
        exit(0);
      }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  while (m--) {
    int x, y;
    cin >> x >> y;
    a[--x].insert(--y);
    a[y].insert(x);
  }
  vector<int> ans(n, -1), groups[3];
  for (int c = 0; c < 3; c++) {
    int found = 0;
    for (int x = 0; x < n; x++)
      if (ans[x] < 0) {
        ans[x] = c;
        groups[c].push_back(x);
        for (int y = 0; y < n; y++)
          if (!a[x].count(y)) {
            if (ans[y] < 0) {
              ans[y] = c;
              groups[c].push_back(y);
            } else if (ans[y] != c) {
              cout << -1 << endl;
              return 0;
            }
          }
        found = 1;
        break;
      }
    if (!found) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++)
    if (ans[i] < 0) {
      cout << -1 << endl;
      return 0;
    }
  for (int i = 0; i < 3; i++)
    for (int x : groups[i])
      if (a[x].size() != n - groups[i].size()) {
        cout << -1 << endl;
        return 0;
      }
  for (int i = 0; i < 3; i++)
    for (int j = i + 1; j < 3; j++) check(groups[i], groups[j]);
  for (int i = 0; i < n; i++) cout << ans[i] + 1 << " \n"[i == n - 1];
}
