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 N, M, K, S, A, B, X, Y, Z, C, a, b, c, d, i, j, k;
void solve() {
  cin >> N >> M;
  vector<long long> V[N];
  map<pair<long long, long long>, bool> Mm;
  for (int i = 0; i < M; i++) {
    cin >> a >> b;
    a--;
    b--;
    V[a].push_back(b);
    V[b].push_back(a);
    Mm[{a, b}] = 1;
    Mm[{b, a}] = 1;
  }
  vector<long long> U[3];
  U[0].push_back(0);
  set<long long> S;
  S.insert(0);
  for (int i = 1; i < N; i++) {
    if (Mm.find({0, i}) == Mm.end()) {
      U[0].push_back(i);
      S.insert(i);
    }
  }
  for (int i = 1; i < N; i++) {
    if (Mm.find({0, i}) != Mm.end()) {
      U[1].push_back(i);
      S.insert(i);
      for (int j = 0; j < N; j++) {
        if (i != j && Mm.find({j, i}) == Mm.end()) {
          U[1].push_back(j);
          if (S.find(j) != S.end()) {
            cout << "-1" << endl;
            return;
          }
          S.insert(j);
        }
      }
      break;
    }
  }
  for (int i = 0; i < N; i++) {
    if (S.find(i) == S.end()) {
      U[2].push_back(i);
      S.insert(i);
      for (int j = 0; j < N; j++) {
        if (i != j && Mm.find({j, i}) == Mm.end()) {
          U[2].push_back(j);
          if (S.find(j) != S.end()) {
            cout << "-1" << endl;
            return;
          }
          S.insert(j);
        }
      }
      break;
    }
  }
  a = U[0].size();
  b = U[1].size();
  c = U[2].size();
  if (b == 0 || c == 0 || a == 0 || (a + b + c != N) ||
      (a * b + b * c + c * a != M)) {
    cout << "-1" << endl;
    return;
  }
  for (int i = 0; i < a; i++) {
    for (int j = 0; j < b; j++) {
      if (Mm.find({U[0][i], U[1][j]}) == Mm.end()) {
        cout << "-1" << endl;
        return;
      }
    }
  }
  for (int i = 0; i < a; i++) {
    for (int j = 0; j < c; j++) {
      if (Mm.find({U[0][i], U[2][j]}) == Mm.end()) {
        cout << "-1" << endl;
        return;
      }
    }
  }
  for (int i = 0; i < c; i++) {
    for (int j = 0; j < b; j++) {
      if (Mm.find({U[2][i], U[1][j]}) == Mm.end()) {
        cout << "-1" << endl;
        return;
      }
    }
  }
  long long ans[N];
  for (int i = 0; i < a; i++) ans[U[0][i]] = 1;
  for (int i = 0; i < b; i++) ans[U[1][i]] = 2;
  for (int i = 0; i < c; i++) ans[U[2][i]] = 3;
  for (int i = 0; i < N; i++) cout << ans[i] << " ";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1, p = 1;
  while (t <= p) {
    solve();
    t++;
  }
  return 0;
}
