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;
long long pow2[200005];
void inithash() {
  long long p = 1;
  for (int i = 0; i < 200002; i++) {
    pow2[i] = p;
    p *= 2;
    p %= 1000000007;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  inithash();
  int n;
  cin >> n;
  vector<vector<int>> ConnectedTo(n);
  int m;
  cin >> m;
  vector<int> ans(n);
  vector<long long> hsh(n);
  vector<int> numVisit(n);
  vector<int> group(n);
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    ConnectedTo[u].push_back(v);
    ConnectedTo[v].push_back(u);
    hsh[u] += pow2[v];
    hsh[v] += pow2[u];
    hsh[u] %= 1000000007;
    hsh[v] %= 1000000007;
  }
  bool can = 1;
  set<long long> poss;
  for (int i = 0; i < n; i++) {
    poss.insert(hsh[i]);
    if (poss.size() > 3) {
      can = 0;
      break;
    }
  }
  if (!can) {
    cout << -1 << "\n";
  } else {
    for (int i = 0; i < n; i++) {
      if (hsh[i] == *poss.begin()) {
        ans[i] = 1;
      } else {
        auto it = poss.end();
        it--;
        if (hsh[i] == *it) {
          ans[i] = 3;
        } else {
          ans[i] = 2;
        }
      }
    }
    long long ha1 = 0;
    long long ha2 = 0;
    long long ha3 = 0;
    for (int i = 0; i < n; i++) {
      if (ans[i] != 1) {
        ha1 += (pow2[i]);
        ha1 %= 1000000007;
      }
      if (ans[i] != 2) {
        ha2 += (pow2[i]);
        ha2 %= 1000000007;
      }
      if (ans[i] != 3) {
        ha3 += (pow2[i]);
        ha3 %= 1000000007;
      }
    }
    if (ha1 != *poss.begin()) {
      can = 0;
    }
    auto it = poss.begin();
    it++;
    auto it2 = poss.end();
    it2--;
    if (ha2 != *it) can = 0;
    if (ha3 != *it2) can = 0;
    if (!can)
      cout << -1 << "\n";
    else {
      for (auto a : ans) cout << a << " ";
      cout << "\n";
    }
  }
  return 0;
}
