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> intersection(set<int> a, set<int> b) {
  set<int> c;
  for (int i : a) {
    if (b.find(i) != b.end()) c.insert(i);
  }
  return c;
}
void solveFunction() {
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > edges;
  set<int> s[n + 1];
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    s[x].insert(y);
    s[y].insert(x);
    edges.push_back({x, y});
  }
  int color[n + 1];
  memset(color, 0, sizeof color);
  for (auto i : edges) {
    if (color[i.first] == 0 && color[i.second] == 0) {
      set<int> inter = intersection(s[i.first], s[i.second]);
      color[i.first] = 1;
      color[i.second] = 2;
      for (auto j : inter) color[j] = 3;
      for (auto j : s[i.first]) {
        if (inter.find(j) == inter.end()) color[j] = 2;
      }
      for (auto j : s[i.second]) {
        if (inter.find(j) == inter.end()) color[j] = 1;
      }
    }
  }
  int cnt[4], edgcnt[4];
  memset(cnt, 0, sizeof cnt);
  memset(edgcnt, 0, sizeof edgcnt);
  for (int i = 1; i <= n; i++) cnt[color[i]]++;
  for (auto i : edges) {
    if (color[i.first] == color[i.second]) {
      cout << -1;
      return;
    }
    edgcnt[color[i.first]]++;
    edgcnt[color[i.second]]++;
  }
  for (int i = 1; i <= 3; i++) {
    if (cnt[i] * (n - cnt[i]) != edgcnt[i] || cnt[i] == 0) {
      cout << -1 << "\n";
      return;
    }
  }
  for (int i = 1; i <= n; i++) cout << color[i] << " ";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    solveFunction();
  }
}
