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 n1 = 1e5 + 5;
vector<int> G[n1];
bool vi[n1];
int cnt;
void dfs(int r) {
  vi[r] = 1;
  cnt++;
  for (int i = 0; i < G[r].size(); ++i) {
    if (vi[G[r][i]] == 0) {
      dfs(G[r][i]);
    }
  }
}
bool ncon(int n) {
  for (int i = 0; i < n1; ++i) {
    vi[i] = 0;
  }
  cnt = 0;
  dfs(1);
  if (cnt == n)
    return false;
  else
    return true;
}
int main() {
  int n, e;
  cin >> n >> e;
  for (int i = 0; i < e; ++i) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  if (ncon(n)) {
    cout << "-1\n";
    return 0;
  }
  set<int> sa;
  for (int i = 0; i < n; ++i) {
    sa.insert(i + 1);
  }
  set<int> sb;
  for (int i = 0; i < G[1].size(); ++i) {
    int r = G[1][i];
    sb.insert(r);
    sa.erase(r);
  }
  if (sb.size() <= 1) {
    cout << "-1";
    return 0;
  }
  int ano = *(sb.begin());
  set<int> sc;
  for (int i = 0; i < G[ano].size(); ++i) {
    int r = G[ano][i];
    if (sb.find(r) != sb.end()) {
      sb.erase(r);
      sc.insert(r);
    }
  }
  if (sc.size() <= 0) {
    cout << "-1\n";
    return 0;
  }
  bool finch = 0;
  set<int>::iterator it = sa.begin();
  int ans[n + 1];
  for (; it != sa.end(); ++it) {
    int r = (*it);
    for (int j = 0; j < G[r].size(); ++j) {
      int r1 = G[r][j];
      if (sa.find(r1) != sa.end()) {
        finch = 1;
        break;
      }
    }
    int uio = sb.size();
    int kl = sc.size();
    uio += kl;
    if (uio != G[r].size()) {
      finch = 1;
      break;
    }
    ans[r] = 1;
  }
  it = sb.begin();
  for (; it != sb.end(); ++it) {
    int r = (*it);
    for (int j = 0; j < G[r].size(); ++j) {
      int r1 = G[r][j];
      if (sb.find(r1) != sb.end()) {
        finch = 1;
        break;
      }
    }
    int uio = sa.size();
    int kl = sc.size();
    uio += kl;
    if (uio != G[r].size()) {
      finch = 1;
      break;
    }
    ans[r] = 2;
  }
  it = sc.begin();
  for (; it != sc.end(); ++it) {
    int r = (*it);
    for (int j = 0; j < G[r].size(); ++j) {
      int r1 = G[r][j];
      if (sc.find(r1) != sc.end()) {
        finch = 1;
        break;
      }
    }
    int uio = sb.size();
    int kl = sa.size();
    uio += kl;
    if (uio != G[r].size()) {
      finch = 1;
      break;
    }
    ans[r] = 3;
  }
  if (finch) {
    cout << "-1\n";
  } else {
    for (int i = 1; i <= n; ++i) {
      cout << ans[i] << " ";
    }
  }
  return 0;
}
