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 overall = 0, n, m, visited[300009];
vector<int> adj[300009];
set<pair<int, int> > s;
int c[4];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    s.insert({u, v});
    s.insert({v, u});
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  visited[1] = 1;
  int flag = 0;
  for (int i = 2; i <= n; i++) {
    if (s.find({1, i}) == s.end()) {
      visited[i] = 1;
    } else if (!flag) {
      flag = i;
      visited[i] = 2;
    }
  }
  int flag2 = 0;
  for (int j = 1; j <= n; j++) {
    if (j == flag) continue;
    if (s.find({flag, j}) == s.end()) {
      if (visited[j] == 1) {
        cout << "-1";
        return 0;
      }
      visited[j] = 2;
    } else if (visited[j] != 1 && !flag2) {
      flag2 = j;
      visited[j] = 3;
    }
  }
  if (!flag2) {
    cout << "-1";
    return 0;
  }
  for (int j = 1; j <= n; j++) {
    if (j == flag2) continue;
    if (s.find({flag2, j}) == s.end()) {
      if (visited[j] == 1 || visited[j] == 2) {
        cout << "-1";
        return 0;
      }
      visited[j] = 3;
    } else {
      if (!visited[j]) {
        cout << "-1";
        return 0;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (visited[i] == 1)
      c[1]++;
    else if (visited[i] == 2)
      c[2]++;
    else
      c[3]++;
  }
  for (int i = 1; i <= n; i++) {
    int cc[4] = {0};
    for (int j = 0; j < adj[i].size(); j++) {
      cc[visited[adj[i][j]]]++;
    }
    if (cc[visited[i]]) {
      cout << "-1";
      return 0;
    }
    for (int j = 1; j <= 3; j++) {
      if (j != visited[i]) {
        if (cc[j] != c[j]) {
          cout << "-1";
          return 0;
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << visited[i] << " ";
  }
  return 0;
}
