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 MAXN = 200009;
const long long MOD = 1e9 + 7;
class {
 public:
  int n, m;
  vector<int> V[MAXN];
  int f[MAXN];
  void solve() {
    cin >> n >> m;
    for (int i = 0; i < m; ++i) {
      int u, v;
      cin >> u >> v;
      V[u].push_back(v);
      V[v].push_back(u);
    }
    vector<int> S[3];
    int j = 0;
    for (int i = 1; i <= n; ++i) {
      if (f[i] == 0) {
        if (j == 3) {
          cout << -1;
          return;
        }
        vector<int> g(n + 9);
        for (int jj : V[i]) g[jj] = 1;
        for (int jj = 1; jj <= n; ++jj)
          if (!g[jj]) S[j].push_back(jj), f[jj] = 1;
        ++j;
      }
    }
    if (j != 3 || S[0].size() + S[1].size() + S[2].size() != n) {
      cout << -1;
      return;
    }
    for (int i = 0; i < 3; ++i) {
      for (int j : S[i]) f[j] = i;
    }
    for (int i = 1; i <= n; ++i) {
      int sz = S[(f[i] + 1) % 3].size() + S[(f[i] + 2) % 3].size();
      if (V[i].size() != sz) {
        cout << -1;
        return;
      }
      for (int j : V[i]) {
        if (f[j] == f[i]) {
          cout << -1;
          return;
        }
      }
    }
    for (int i = 1; i <= n; ++i) {
      cout << f[i] + 1 << " ";
    }
  }
} NSPACE;
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  NSPACE.solve();
}
