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;
using ll = long long;
using ld = long double;
const int maxn = 100500;
const int inf = 1e9 + 5;
const int mod = 1e9 + 7;
int n, m;
vector<int> g[maxn];
set<int> used;
int num[maxn];
int cur_num = 0;
void dfs(int i) {
  used.erase(i);
  num[i] = cur_num;
  int last = 0;
  g[i].push_back(n + 1);
  sort((g[i]).begin(), (g[i]).end());
  for (int j : g[i]) {
    int l = last + 1;
    int r = j - 1;
    while (!used.empty()) {
      auto it = used.lower_bound(l);
      if (it == used.end() || *it > r) {
        break;
      }
      dfs(*it);
    }
    last = j;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int i = 1; i <= n; ++i) {
    used.insert(i);
  }
  while (!used.empty()) {
    cur_num++;
    dfs(*used.begin());
  }
  if (cur_num != 3) {
    cout << -1 << endl;
    exit(0);
  }
  for (int i = 1; i <= n; ++i) {
    for (int x : g[i])
      if (x != n + 1 && num[x] == num[i]) {
        cout << -1 << endl;
        return 0;
      }
  }
  for (int i = 1; i <= n; ++i) {
    cout << num[i] << " \n"[i == n];
  }
}
