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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int N, M;
int color[100001];
vector<int> adj[100001];
void dfs(int node) {
  for (int x : adj[node]) {
    if (color[x] == 4) {
      if (color[node] == 2)
        color[x] = 3;
      else
        color[x] = 2;
      dfs(node);
    }
  }
}
vector<int> c[4];
map<pair<int, int>, bool> edge;
void fail() {
  cout << -1 << endl;
  exit(0);
}
void check() {
  long long edges = 0;
  for (int i = 1; i <= 3; i++)
    for (int j = i + 1; j <= 3; j++)
      edges += ((long long)(c[i].size())) * ((long long)(c[j].size()));
  if (edges != M) fail();
  for (int i = 1; i <= 3; i++)
    if (c[i].size() == 0) fail();
  for (int i = 1; i <= 3; i++) {
    for (int j = i + 1; j <= 3; j++) {
      for (int x : c[i])
        for (int y : c[j])
          if (!edge[{x, y}]) fail();
    }
  }
}
int main() {
  scanf("%d%d", &N, &M);
  for (int i = 0; i < M; i++) {
    int X, Y;
    scanf("%d%d", &X, &Y);
    edge[{X, Y}] = true;
    edge[{Y, X}] = true;
    adj[X].push_back(Y);
    adj[Y].push_back(X);
  }
  color[1] = 1;
  for (int x : adj[1]) color[x] = 4;
  for (int i = 1; i <= N; i++)
    if (color[i] == 0) color[i] = 1;
  for (int x : adj[1])
    if (color[x] == 4) {
      color[x] = 2;
      dfs(x);
    }
  for (int i = 1; i <= N; i++) c[color[i]].push_back(i);
  check();
  for (int i = 1; i <= N; i++) cout << color[i] << ' ';
  cout << endl;
  return 0;
}
