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;
map<pair<int, int>, int> edge;
vector<pair<int, int> > cac;
int e[300005];
int n, m;
int color[300005];
int dem[50][50];
int main() {
  cin >> n >> m;
  if (m == 0) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    edge[pair<int, int>(u, v)] = 1;
    edge[pair<int, int>(v, u)] = 1;
    cac.push_back(pair<int, int>(u, v));
    e[u] = 1;
    e[v] = 1;
  }
  int id = 0;
  color[1] = 1;
  for (int i = 1; i <= n; i++) {
    if (color[i]) continue;
    if (edge[pair<int, int>(1, i)]) continue;
    color[i] = 1;
  }
  int check = 0;
  for (int i = 1; i <= n; i++) check += e[i];
  for (int i = 1; i <= n; i++) {
    if (color[i]) continue;
    color[i] = 2;
    for (int j = i; j <= n; j++) {
      if (color[j]) continue;
      if (edge[pair<int, int>(i, j)]) continue;
      color[j] = 2;
    }
    break;
  }
  for (int i = 1; i <= n; i++) {
    if (color[i]) continue;
    color[i] = 3;
    for (int j = i; j <= n; j++) {
      if (color[j]) continue;
      if (edge[pair<int, int>(i, j)]) continue;
      color[j] = 3;
    }
    break;
  }
  for (int i = 0; i < m; i++) {
    int u = cac[i].first;
    int v = cac[i].second;
    dem[color[u]][color[v]]++;
    dem[color[v]][color[u]]++;
  }
  pair<int, int> x;
  int cnt[5];
  cnt[1] = cnt[2] = cnt[3] = 0;
  long long ans = 0;
  for (int i = 1; i <= n; i++) cnt[color[i]]++;
  int cacac = 0;
  for (int i = 1; i <= 3; i++)
    for (int j = i + 1; j <= 3; j++) {
      if (1LL * cnt[i] * 1LL * cnt[j] != 1LL * dem[i][j]) cacac = 1;
    }
  ans = 1LL * cnt[1] * cnt[2] + 1LL * cnt[2] * cnt[3] + 1LL * cnt[3] * cnt[1];
  if (ans != 1LL * m || cnt[1] + cnt[2] + cnt[3] != n ||
      !(cnt[1] && cnt[2] && cnt[3]) || check != n || cacac) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i <= n; i++) cout << color[i] << " ";
}
