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 long long INF = 1e9;
void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
vector<long long> vc[100001];
long long vst[300001];
int32_t main() {
  fastio();
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    long long a, b;
    cin >> a >> b;
    vc[a].push_back(b);
    vc[b].push_back(a);
  }
  vector<long long> v1, v2, v3;
  long long st = 0, fg = 0;
  for (long long i = 0; i < vc[1].size(); i++) {
    vst[vc[1][i]] = 2;
    st = vc[1][i];
  }
  for (long long i = 1; i < n + 1; i++) {
    if (!vst[i]) {
      vst[i] = 1;
      v1.push_back(i);
    }
  }
  if (st == 0) fg = 1;
  for (long long i = 0; i < vc[st].size(); i++) {
    if (vst[vc[st][i]] == 2) {
      vst[vc[st][i]] = 3;
      v3.push_back(vc[st][i]);
    }
  }
  long long s2 = n - v1.size() - v3.size();
  if (s2 == 0 || v1.size() == 0 || v3.size() == 0) fg = 1;
  for (long long i = 1; i < n + 1; i++) {
    long long sz;
    if (vst[i] == 1)
      sz = n - v1.size();
    else if (vst[i] == 2)
      sz = n - s2;
    else
      sz = n - v3.size();
    if (vc[i].size() < sz) fg = 1;
    for (long long j = 0; j < vc[i].size(); j++) {
      if (vst[vc[i][j]] == vst[i]) fg = 1;
    }
  }
  if (fg == 1)
    cout << -1;
  else {
    for (long long i = 1; i < n + 1; i++) cout << vst[i] << " ";
  }
}
