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;
const int N = 1e5 + 5, mod = 1e9 + 7, mod1 = 998242353, mod2 = 1e9 + 9,
          inf = 1e9 + 7;
const long long infll = 1e18 + 7;
int n, m;
set<int> adj[N];
int par[N];
int col[N];
bool ck[N];
vector<int> col1, col2, col3;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(par, -1, sizeof(par));
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].insert(y);
    adj[y].insert(x);
  }
  for (auto v = adj[1].begin(); v != adj[1].end(); v++) {
    ck[*v] = 1;
  }
  col[1] = 1;
  col1.push_back(1);
  for (int i = 1; i <= n; i++) {
    if (!ck[i]) {
      col[i] = 1;
      col1.push_back(i);
    }
  }
  int idx = 0;
  for (int i = 1; i <= n; i++) {
    if (col[i] == 0) {
      idx = i;
      break;
    }
  }
  memset(ck, 0, sizeof(ck));
  for (auto v = adj[idx].begin(); v != adj[idx].end(); v++) {
    if (col[*v] == 0) {
      ck[*v] = 1;
    }
  }
  col[idx] = 2;
  col2.push_back(idx);
  for (int i = 1; i <= n; i++) {
    if (!ck[i] && col[i] == 0) {
      col[i] = 2;
      col2.push_back(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (col[i] == 0) {
      col[i] = 3;
      col3.push_back(i);
    }
  }
  if (col3.size() == 0) return cout << -1, 0;
  ;
  for (int i = 1; i <= n; i++) {
    for (auto v = adj[i].begin(); v != adj[i].end(); v++) {
      if (col[i] == col[*v]) return cout << -1, 0;
      ;
    }
  }
  for (auto u : col1) {
    for (auto v : col2) {
      if (adj[u].find(v) == adj[u].end()) return cout << -1, 0;
      ;
      if (adj[v].find(u) == adj[v].end()) return cout << -1, 0;
      ;
    }
  }
  for (auto u : col2) {
    for (auto v : col3) {
      if (adj[u].find(v) == adj[u].end()) return cout << -1, 0;
      ;
      if (adj[v].find(u) == adj[v].end()) return cout << -1, 0;
      ;
    }
  }
  for (auto u : col3) {
    for (auto v : col1) {
      if (adj[u].find(v) == adj[u].end()) return cout << -1, 0;
      ;
      if (adj[v].find(u) == adj[v].end()) return cout << -1, 0;
      ;
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << col[i] << ' ';
  }
}
