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 mod = 998244353;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, m, check[100001], cnt = 0, ans = 1;
  cin >> n >> m;
  vector<int> node[n + 1], v;
  while (m--) {
    int x, y;
    cin >> x >> y;
    node[x].push_back(y);
    node[y].push_back(x);
  }
  for (long long i = 1; i <= n; i++) {
    long long hash = 0;
    for (long long i2 : node[i]) {
      hash = (hash + i2 * i2 * i2) % mod;
    }
    check[i] = hash;
    if (!count(v.begin(), v.end(), hash)) {
      v.push_back(hash);
      cnt += node[i].size();
    }
    if (v.size() > 3) {
      ans = 0;
      break;
    }
  }
  if (cnt != 2 * n) ans = 0;
  if (ans == 0 or v.size() != 3)
    cout << -1;
  else {
    for (long long i = 1; i <= n; i++) {
      for (long long i2 = 0; i2 < 3; i2++) {
        if (check[i] == v[i2]) cout << i2 + 1 << " ";
      }
    }
  }
}
