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>
#pragma GCC optimize("O3")
#pragma GCC target("avx2")
#pragma GCC target(sse, sse2, sse3, ssse3, sse4, popcnt, tune = native)
using namespace std;
const long long MAXN = 1e5 + 47;
long long n, m;
vector<vector<long long> > graf(MAXN);
void read() {
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    long long a, b;
    cin >> a >> b;
    --a, --b;
    graf[a].push_back(b);
    graf[b].push_back(a);
  }
}
void bad() {
  cout << -1 << endl;
  exit(0);
}
vector<long long> color(MAXN, 0);
void dfs(long long p1) {
  color[p1] = 1;
  for (long long p2 : graf[p1]) {
    if (color[p2] == 1) continue;
    dfs(p2);
  }
}
void solve() {
  color.assign(MAXN, 0);
  dfs(0);
  long long C = 0;
  for (long long i = 0; i < n; i++)
    if (color[i] == 1) C++;
  if (n <= 2 || C != n) bad();
  set<long long> white;
  for (long long i = 0; i < n; i++) white.insert(i);
  for (long long p2 : graf[0]) {
    white.erase(p2);
  }
  vector<long long> cnt(MAXN, 0);
  for (long long p : white) {
    for (long long ch : graf[p]) cnt[ch]++;
  }
  long long szw = (long long)(white.size()), szb = 0, szg = n - szw;
  long long start = -1;
  for (long long i = 0; i < MAXN; i++) {
    if (cnt[i] == szw) szb++, start = i;
    if (cnt[i] != 0 && cnt[i] != szw) {
      cout << -1 << endl;
      return;
    }
  }
  if (szg != szb) {
    cout << -1 << endl;
    return;
  }
  set<long long> black;
  for (long long i = 0; i < n; i++) black.insert(i);
  for (long long p2 : graf[start]) {
    black.erase(p2);
  }
  cnt.assign(MAXN, 0);
  for (long long p : black) {
    for (long long ch : graf[p]) cnt[ch]++;
  }
  szb = (long long)(black.size()), szw = 0, szg = n - szb;
  for (long long i = 0; i < MAXN; i++) {
    if (cnt[i] == szb) szw++;
    if (cnt[i] != 0 && cnt[i] != szb) {
      cout << -1 << endl;
      return;
    }
  }
  if (szw != szg) {
    cout << -1 << endl;
    return;
  }
  vector<long long> ans(n, 0);
  for (long long p : white) ans[p] = 1;
  for (long long p : black)
    if (ans[p] != 0) {
      cout << -1 << endl;
    } else {
      ans[p] = 2;
    }
  start = -1;
  for (long long i = 0; i < n; i++)
    if (ans[i] == 0) start = i;
  if (start == -1) bad();
  set<long long> green;
  for (long long i = 0; i < n; i++) green.insert(i);
  for (long long p : white) green.erase(p);
  for (long long p : black) green.erase(p);
  cnt.assign(MAXN, 0);
  for (long long p : green) {
    for (long long ch : graf[p]) cnt[ch]++;
  }
  szg = (long long)(green.size()), szw = 0, szb = n - szg;
  for (long long i = 0; i < MAXN; i++) {
    if (cnt[i] == szg) szw++;
    if (cnt[i] != 0 && cnt[i] != szg) {
      cout << -1 << endl;
      return;
    }
  }
  if (szw != szb) {
    cout << -1 << endl;
    return;
  }
  for (long long p : green) {
    if (ans[p] != 0) bad();
    ans[p] = 3;
  }
  for (long long i = 0; i < n; i++)
    if (ans[i] == 0) bad();
  for (long long p : ans) cout << p << " ";
  cout << endl;
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  long long t = 1;
  while (t--) {
    read();
    solve();
  }
  return 0;
}
