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() {
  int n, m;
  scanf("%d %d", &n, &m);
  vector<set<int>> adj(n);
  set<pair<int, int>> st;
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--, v--;
    adj[u].insert(v);
    adj[v].insert(u);
    st.insert({u, v});
  }
  vector<int> a(n, -1);
  a[0] = 1;
  vector<int> v1(n + 5);
  int ind = 0;
  v1[ind++] = 0;
  int l = -1;
  for (int i = 1; i < n; i++) {
    if (adj[0].find(i) == adj[0].end()) {
      a[i] = 1;
      v1[ind++] = i;
    } else {
      l = i;
    }
  }
  v1.resize(ind);
  if (l == -1) {
    cout << -1 << endl;
    return 0;
  }
  vector<int> v2(n + 5);
  ind = 0;
  v2[ind++] = l;
  a[l] = 2;
  for (int i = 1; i < n; i++) {
    if (i == l) continue;
    if (a[i] == -1 && adj[l].find(i) == adj[l].end()) {
      a[i] = 2;
      v2[ind++] = i;
    }
  }
  v2.resize(ind);
  ind = 0;
  vector<int> v3(n + 5);
  for (int i = 1; i < n; i++) {
    if (a[i] == -1) {
      a[i] = 3;
      v3[ind++] = i;
    }
  }
  v3.resize(ind);
  int s1, s2, s3;
  s1 = v1.size(), s2 = v2.size(), s3 = v3.size();
  if (s1 == 0 || s2 == 0 || s3 == 0) {
    cout << -1 << endl;
    return 0;
  }
  for (auto k : st) {
    int u = k.first, v = k.second;
    if (a[u] == a[v]) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    if (a[i] == 1) {
      int k = (int)adj[i].size();
      if (k != (s2 + s3)) {
        cout << -1 << endl;
        return 0;
      }
    } else if (a[i] == 2) {
      int k = (int)adj[i].size();
      if (k != (s1 + s3)) {
        cout << -1 << endl;
        return 0;
      }
    } else if (a[i] == 3) {
      int k = (int)adj[i].size();
      if (k != (s1 + s2)) {
        cout << -1 << endl;
        return 0;
      }
    }
  }
  for (int i : a) {
    cout << i << " ";
  }
  cout << endl;
}
