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;
long long F[1000005], used[1000005];
vector<pair<long long, long long> > E;
map<pair<long long, long long>, long long> Ed;
signed main() {
  ios_base::sync_with_stdio(false);
  long long t, j = 0, k = 0, y, g, sum = 0, i, q, l, n, m, ans = 1, x;
  long long u, v, x1 = 0, x2 = 0, x3 = 0, flag = 1;
  set<long long> s;
  cin >> n >> m;
  while (m--) {
    cin >> u >> v;
    F[u]++;
    F[v]++;
    E.push_back({u, v});
    Ed[{u, v}] = Ed[{v, u}] = 1;
  }
  for (i = 1; i <= n; i++) {
    if (Ed[{1, i}] == 0) {
      used[i] = 1;
      x1++;
    }
  }
  for (i = 1; i <= n; i++) {
    if (used[i] == 0) {
      j = i;
      break;
    }
  }
  for (i = 1; i <= n; i++) {
    if (Ed[{j, i}] == 0 && j > 0) {
      if (used[i] != 0) flag = 0;
      x2++;
      used[i] = 2;
    }
  }
  for (i = 1; i <= n; i++) {
    if (used[i] == 0) {
      k = i;
      break;
    }
  }
  for (i = 1; i <= n; i++) {
    if (Ed[{k, i}] == 0 && k > 0) {
      if (used[i] != 0) flag = 0;
      x3++;
      used[i] = 3;
    }
  }
  if (x1 == 0 || x2 == 0 || x3 == 0) flag = 0;
  for (auto e : E) {
    u = e.first;
    v = e.second;
    if (used[u] == used[v]) flag = 0;
  }
  for (i = 1; i <= n; i++) {
    if (used[i] == 0) flag = 0;
    if (used[i] == 1 && F[i] != x2 + x3) flag = 0;
    if (used[i] == 2 && F[i] != x1 + x3) flag = 0;
    if (used[i] == 3 && F[i] != x2 + x1) flag = 0;
  }
  if (flag == 0) {
    cout << -1;
    return 0;
  }
  for (i = 1; i <= n; i++) cout << used[i] << " ";
  return 0;
}
