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;
typedef int (*uniopi)(int);
typedef long long (*uniopl)(long long);
typedef int (*binopi)(int, int);
typedef long long (*binopl)(long long, long long);
typedef int (*tupopi)(int, int, int);
typedef long long (*tupopl)(long long, long long, long long);
const long double PI = acos(0) * 2;
const int INF = 0x3f2f1f0f;
const long long LINF = 1ll * INF * INF;
int X4[] = {-1, 0, 1, 0}, Y4[] = {0, -1, 0, 1};
int X8[] = {-1, -1, -1, 0, 1, 1, 1, 0}, Y8[] = {-1, 0, 1, 1, 1, 0, -1, -1};
template <typename T>
inline bool uax(T &a, T b) {
  return a < b ? (a = b, 1) : 0;
}
template <typename T>
inline bool uin(T &a, T b) {
  return a > b ? (a = b, 1) : 0;
}
template <typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
  return is >> p.first >> p.second;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  return os << '(' << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (int i = 0; i < (int)v.size(); i++) {
    os << v[i];
    if (i + 1 < (int)v.size()) os << ' ';
  }
  return os;
}
const int MAX_N = 1e5 + 10;
int N;
vector<int> Ed[MAX_N];
vector<pair<int, int> > Eds;
int Vst[MAX_N];
int isTripartite(vector<int> &a, vector<int> &b, vector<int> &c) {
  int na = ((int)(a).size()), nb = ((int)(b).size()), nc = ((int)(c).size());
  if (((int)(a).size()) + ((int)(b).size()) + ((int)(c).size()) != N ||
      na == 0 || nb == 0 || nc == 0)
    return 0;
  vector<int> s(N + 1, 0);
  for (auto &x : a) {
    if (s[x]) return 0;
    s[x] = 1;
    if (((int)(Ed[x]).size()) != nb + nc) return 0;
  }
  for (auto &x : b) {
    if (s[x]) return 0;
    s[x] = 2;
    if (((int)(Ed[x]).size()) != nc + na) return 0;
  }
  for (auto &x : c) {
    if (s[x]) return 0;
    s[x] = 3;
    if (((int)(Ed[x]).size()) != na + nb) return 0;
  }
  for (auto &p : Eds) {
    int x, y;
    tie(x, y) = p;
    if (s[x] == s[y]) return 0;
  }
  return 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cout << std::setprecision(10);
  cout << fixed;
  int m;
  cin >> N >> m;
  for (int __i = 0; __i < (m); __i++) {
    int a, b;
    cin >> a >> b;
    Ed[a].push_back(b);
    Ed[b].push_back(a);
    Eds.emplace_back(a, b);
  }
  vector<int> a, nota, ina(N + 1, 1);
  for (auto &x : Ed[1]) ina[x] = 0;
  for (int i = 1; i <= (N); i++) {
    if (ina[i])
      a.push_back(i);
    else
      nota.push_back(i);
  }
  if (((int)(nota).size()) == 0) {
    cout << -1 << endl;
    return 0;
  }
  vector<int> b, notb, inb(N + 1, 1), c;
  for (auto &x : Ed[nota[0]]) inb[x] = 0;
  for (int i = 1; i <= (N); i++) {
    if (inb[i])
      b.push_back(i);
    else if (!ina[i])
      c.push_back(i);
  }
  if (isTripartite(a, b, c)) {
    vector<int> ans(N + 1);
    for (auto x : a) ans[x] = 1;
    for (auto x : b) ans[x] = 2;
    for (auto x : c) ans[x] = 3;
    for (int i = 1; i <= (N); i++) cout << ans[i] << ' ';
    cout << endl;
  } else {
    cout << -1 << endl;
  }
  return 0;
}
