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 mod(long long int x) { return x % 1000000007; }
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int mod(long long int num, long long int den) {
  return mod(mod(num) * power(den, 1000000007 - 2, 1000000007));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, m;
  cin >> n >> m;
  vector<pair<long long int, long long int> > ed;
  for (long long int i = 0; i < m; ++i) {
    long long int a, b;
    cin >> a >> b;
    ed.push_back({min(a, b), max(a, b)});
  }
  sort(ed.begin(), ed.end());
  vector<long long int> grp(n + 1);
  long long int ma = 0;
  for (long long int i = 0; i < m; ++i) {
    auto a = ed[i].first;
    auto b = ed[i].second;
    if (grp[a] == grp[b]) {
      grp[b] = grp[a] + 1;
      ma = max(ma, grp[a]);
      ma = max(ma, grp[b]);
    }
  }
  bool ok = true;
  if (ma != 2) ok = false;
  long long int one, two, three;
  one = two = three = 0;
  for (long long int i = 1; i <= n; ++i) {
    if (grp[i] == 0)
      ++one;
    else if (grp[i] == 1)
      ++two;
    else
      ++three;
  }
  long long int v12, v13, v23;
  v12 = v13 = v23 = 0;
  for (long long int i = 0; i < m; ++i) {
    auto a = ed[i].first;
    auto b = ed[i].second;
    if (grp[a] == 0) {
      if (grp[b] == 1)
        v12++;
      else if (grp[b] == 2)
        v13++;
    } else if (grp[a] == 1) {
      if (grp[b] == 0)
        v12++;
      else if (grp[b] == 2)
        v23++;
    } else {
      if (grp[b] == 0)
        v13++;
      else
        v23++;
    }
  }
  if (v12 != one * two) ok = false;
  if (v13 != one * three) ok = false;
  if (v23 != two * three) ok = false;
  if (!ok)
    cout << -1 << endl;
  else {
    for (long long int i = 1; i <= n; ++i) {
      cout << grp[i] + 1 << " ";
    }
    cout << endl;
  }
}
