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;
set<int> a[101000];
int ans[101000];
bool used[101000];
int b[5], k[5];
bool f = true;
void dfs(int v, int mn) {
  used[v] = true;
  ans[v] = mn;
  b[mn] = v;
  k[mn]++;
  for (int i : a[v])
    if (!used[i]) {
      for (int j = 1; j <= 3; j++)
        if (j != mn && !a[i].count(b[j])) {
          dfs(i, j);
          break;
        }
      if (!used[i]) {
        used[i] = true;
        f = false;
      }
    } else if (ans[i] == mn)
      f = false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int i, n, m, bb, c;
  cin >> n >> m;
  b[1] = b[2] = b[3] = -1;
  for (i = 0; i < m; i++) {
    cin >> bb >> c;
    bb--;
    c--;
    a[bb].insert(c);
    a[c].insert(bb);
  }
  dfs(0, 1);
  if (k[1] == 0 || k[2] == 0 || k[3] == 0) f = false;
  for (i = 0; i < n; i++)
    if (!used[i] || ans[i] == 0 ||
        a[i].size() != k[1] + k[2] + k[3] - k[ans[i]])
      f = false;
  if (f)
    for (i = 0; i < n; i++) cout << ans[i] << ' ';
  else
    cout << -1;
  return 0;
}
