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;
const int maxn = 1e5 + 7;
vector<int> vec[maxn], ve[10];
int pre[maxn];
bool vis[maxn];
bool dfs(int u) {
  vis[u] = true;
  for (auto v : vec[u]) {
    if (pre[v] == pre[u]) {
      return false;
    }
    if (vis[v] == false) {
      if (dfs(v) == false) {
        return false;
      }
    }
  }
  return true;
}
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    vec[u].push_back(v);
    vec[v].push_back(u);
  }
  for (auto v : vec[1]) {
    pre[v] = 1;
  }
  int t = 1;
  for (int i = 1; i <= n; i++) {
    if (pre[i] == 1) {
      t = i;
      break;
    }
  }
  for (auto v : vec[t]) {
    if (pre[v] == 1) {
      pre[v] = 2;
    }
  }
  for (int i = 1; i <= n; i++) {
    ve[pre[i]].push_back(i);
  }
  bool ok = true;
  for (int i = 0; i < 3; i++) {
    if (ok == false) break;
    for (auto x : ve[i]) {
      if (vec[x].size() != n - ve[i].size()) {
        ok = false;
        break;
      }
    }
  }
  if (ve[2].size() == 0) {
    ok = false;
  }
  for (int i = 1; i <= n; i++) {
    if (vis[i] == false) {
      if (dfs(i) == false) {
        ok = false;
        break;
      }
    }
    if (ok == false) {
      break;
    }
  }
  if (ok) {
    for (int i = 1; i <= n; i++) {
      cout << pre[i] + 1 << ' ';
    }
    cout << endl;
  } else {
    puts("-1");
  }
  return 0;
}
