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 long long N = 100005;
char en = '\n';
long long inf = 1e16;
long long mod = 1e9 + 7;
long long power(long long x, long long n, long long mod) {
  long long res = 1;
  x %= mod;
  while (n) {
    if (n & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    n >>= 1;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n >> m;
  map<pair<long long, long long>, long long> edges;
  map<long long, long long> adj[n + 5];
  for (long long i = 1; i <= m; i++) {
    long long a, b;
    cin >> a >> b;
    if (a > b) swap(a, b);
    adj[a][b];
    adj[b][a];
    edges[{a, b}];
  }
  bool visited[n + 5];
  memset(visited, false, sizeof(visited));
  vector<long long> arr[3];
  arr[0].push_back(1);
  visited[1] = true;
  for (long long i = 1; i <= n; i++) {
    if (i == 1) continue;
    if (!adj[1].count(i)) {
      visited[i] = true;
      arr[0].push_back(i);
    }
  }
  long long i;
  for (i = 1; i <= n; i++) {
    if (!visited[i]) {
      break;
    }
  }
  if (i == n + 1) {
    cout << -1 << en;
    return 0;
  }
  visited[i] = true;
  arr[1].push_back(i);
  for (long long j = 1; j <= n; j++) {
    if (j == i) continue;
    if (visited[j] or adj[i].count(j)) continue;
    arr[1].push_back(j);
    visited[j] = true;
  }
  for (i = 1; i <= n; i++) {
    if (!visited[i]) arr[2].push_back(i);
  }
  long long cnt2 = 0;
  for (long long i = 0; i <= 2; i++) {
    for (long long j = i + 1; j <= 2; j++) {
      cnt2 += (long long)arr[i].size() * arr[j].size();
    }
  }
  if (cnt2 != m) {
    cout << -1 << en;
    return 0;
  }
  long long ans[n + 5];
  for (long long i = 0; i <= 2; i++) {
    if (arr[i].size() == 0) {
      cout << -1 << en;
      return 0;
    }
    for (long long x : arr[i]) ans[x] = i + 1;
    for (long long j = i + 1; j <= 2; j++) {
      for (long long &x : arr[i]) {
        for (long long &y : arr[j]) {
          if (!edges.count({min(x, y), max(x, y)})) {
            cout << -1 << en;
            return 0;
          }
        }
      }
    }
  }
  for (long long i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << en;
  return 0;
}
