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;
using namespace std::chrono;
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
mt19937 rng(steady_clock::now().time_since_epoch().count());
void usaco(string filename) {
  freopen((filename + ".in").c_str(), "r", stdin);
  freopen((filename + ".out").c_str(), "w", stdout);
}
const long long mod = 1000000007;
long long madd(long long a, long long b) { return (a + b) % mod; }
long long msub(long long a, long long b) {
  return (((a - b) % mod) + mod) % mod;
}
long long mmul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long mpow(long long base, long long exp) {
  long long res = 1;
  while (exp) {
    if (exp % 2 == 1) {
      res = (res * base) % mod;
    }
    exp >>= 1;
    base = (base * base) % mod;
  }
  return res;
}
long long minv(long long base) { return mpow(base, mod - 1); }
long long mdiv(long long a, long long b) { return mmul(a, minv(b)); }
long long gcd(long long x, long long y) {
  if (x == 0) return y;
  if (y == 0) return x;
  return gcd(y, x % y);
}
bool prime[1000006];
void sieve(int n) {
  for (long long i = 0; i < n + 1; i++) prime[i] = 1;
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  prime[1] = prime[0] = 0;
}
bool isPrime(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long n, m, q, Q, T, k, l, r, x, y, z, g;
long long a[500005];
long long b[500005];
string second, t;
long long ans = 0;
vector<int> edges[100005];
int col[100005];
bool ok[100005][3];
bool vis[100005];
int sz[3];
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
  };
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    edges[a].push_back(b);
    edges[b].push_back(a);
  };
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < 3; j++) ok[i][j] = 1;
  for (long long i = 0; i < n; i++) col[i] = -1;
  ok[0][1] = 0;
  ok[0][2] = 0;
  memset(vis, 0, sizeof(vis));
  memset(sz, 0, sizeof(sz));
  bool F = 1;
  queue<int> q;
  q.push(0);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    if (vis[x]) continue;
    vis[x] = 1;
    for (long long i = 0; i < 3; i++) {
      if (ok[x][i]) {
        col[x] = i;
        break;
      }
    }
    if (col[x] == -1) {
      F = 0;
      break;
    }
    for (int i : edges[x]) {
      ok[i][col[x]] = 0;
      if (!vis[i]) q.push(i);
    }
  }
  for (long long i = 0; i < n; i++) sz[col[i]]++;
  for (long long i = 0; i < n; i++)
    if (edges[i].size() != n - sz[col[i]]) F = 0;
  for (long long i = 0; i < n; i++)
    if (col[i] == -1) F = 0;
  for (long long i = 0; i < 3; i++)
    if (sz[i] == 0) F = 0;
  if (!F) {
    cout << -1 << endl;
  } else {
    for (long long i = 0; i < n; i++) cout << col[i] + 1 << " ";
  }
}
