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 int powi(long long int a, long long int b) {
  if (b == 0) return 1;
  long long int see = powi(a, b / 2);
  see *= see;
  see %= 1000000007;
  if (b % 2) see *= a;
  see %= 1000000007;
  return see;
}
long long int n, m, x, y;
vector<long long int> v[500009];
long long int color[500009];
long long int ones = 0;
long long int twos = 0;
long long int threes = 0;
void pre() {}
bool check() {
  long long int one = 0, two = 0, three = 0;
  for (long long int i = 0; i < n; i++) {
    one = 0;
    two = 0;
    three = 0;
    for (auto j : v[i]) {
      if (color[j] == 1) {
        one++;
      } else if (color[j] == 2) {
        two++;
      } else {
        three++;
      }
    }
    if (color[i] == 1) {
      if (one == 0 && two == twos && three == threes) continue;
    }
    if (color[i] == 2) {
      if (one == ones && two == 0 && three == threes) continue;
    }
    if (color[i] == 3) {
      if (one == ones && two == twos && three == 0) continue;
    }
    return 0;
  }
  return 1;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  for (long long int i = 0; i < n; i++) {
    color[i] = 0;
  }
  for (long long int i = 0; i < m; i++) {
    cin >> x >> y;
    x--;
    y--;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (auto i : v[0]) {
    color[i] = -1;
  }
  long long int nxt = -1;
  for (long long int i = 0; i < n; i++) {
    if (color[i] == 0) {
      color[i] = 1;
      ones++;
    } else {
      nxt = i;
    }
  }
  if (nxt == -1) {
    cout << "-1\n";
    return 0;
  }
  for (auto i : v[nxt]) {
    if (color[i] == 1) continue;
    color[i] = 3;
    threes++;
  }
  nxt = -1;
  for (long long int i = 0; i < n; i++) {
    if (color[i] == -1) {
      color[i] = 2;
      twos++;
    } else if (color[i] == 3) {
      nxt = i;
    }
  }
  if (nxt == -1) {
    cout << "-1\n";
    return 0;
  }
  if (check() == 0 || ones == 0 || twos == 0 || threes == 0) {
    cout << "-1\n";
    return 0;
  }
  for (long long int i = 0; i < n; i++) {
    cout << color[i] << " ";
  }
  cout << "\n";
}
