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 long long int logx = 30;
const long long int N = 1e5 + 5;
const long long int M = 1e3 + 5;
const long long int mod = 1e9 + 7;
const long long int INF = 1e18 + 5;
const double PI = 3.14159265358979323846;
inline long long int mul(long long int a, long long int b) {
  return (a * 1ll * b) % mod;
}
inline long long int sub(long long int a, long long int b) {
  long long int c = a - b;
  if (c < 0) c += mod;
  return c;
}
inline long long int add(long long int a, long long int b) {
  long long int c = a + b;
  if (c > mod) c -= mod;
  return c;
}
template <typename T>
T power(T x, T y, long long int m) {
  T ans = 1;
  while (y > 0) {
    if (y & 1LL) ans = (ans * x) % m;
    y >>= 1ll;
    x = (x * x) % m;
  }
  return ans % m;
}
set<long long int> second;
vector<long long int> g[N];
long long int n, m, cnt, col[N];
void BFS() {
  cnt++;
  long long int cur = *second.begin();
  second.erase(second.begin());
  queue<long long int> q;
  q.push(cur);
  while (q.size()) {
    vector<long long int> temp;
    cur = q.front();
    q.pop();
    col[cur] = cnt;
    for (auto i : second) {
      long long int lb =
          lower_bound(g[cur].begin(), g[cur].end(), i) - g[cur].begin();
      if (lb == g[cur].size()) {
        q.push(i);
        temp.push_back(i);
      } else {
        if (g[cur][lb] == i) {
          continue;
        }
        q.push(i);
        temp.push_back(i);
      }
    }
    for (auto i : temp) {
      second.erase(i);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  cin >> n >> m;
  for (long long int i = 1; i <= n; i++) {
    second.insert(i);
  }
  while (m--) {
    long long int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (long long int i = 1; i <= n; i++) {
    sort(g[i].begin(), g[i].end());
  }
  while (second.size()) {
    BFS();
  }
  if (cnt != 3) {
    cout << -1 << endl;
    exit(0);
  }
  for (long long int i = 1; i <= n; i++) {
    for (auto j : g[i]) {
      if (col[i] == col[j]) {
        cout << -1 << endl;
        exit(0);
      }
    }
  }
  for (long long int i = 1; i <= n; i++) {
    cout << col[i] << " ";
  }
}
