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;
long long power(long long x, long long y) {
  if (y == 0) {
    return 1;
  }
  long long p = power(x, y / 2);
  p = (p * p);
  return (y % 2 == 0) ? p : (x * p);
}
long long powm(long long x, long long y, long long m) {
  long long ans = 1, r = 1;
  x %= m;
  while (r > 0 && r <= y) {
    if (r & y) {
      ans *= x;
      ans %= m;
    }
    r <<= 1;
    x *= x;
    x %= m;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) {
  long long lar = max(a, b);
  long long small = min(a, b);
  for (long long i = lar;; i += lar) {
    if (i % small == 0) {
      return i;
    }
  }
}
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
bool sortbydec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.first > b.first);
}
void dex(int a) {
  if (a == 1) {
    cout << "YES";
  }
  if (a == 2) {
    cout << "NO";
  }
  if (a == 3) {
    cout << "Yes";
  }
  if (a == 4) {
    cout << "No";
  }
}
void line(long long x) {
  for (long long i = 1; i <= x; i++) {
    cout << "-";
  }
  cout << '\n';
}
const long long N = (long long)(300005);
const long long MOD = (long long)(1e9 + 7);
const long long INF = (long long)(2e18);
long long pw[N];
long long ha[N];
map<long long, long long> ma;
long long n, m;
bool vis[N];
vector<long long> adj[N];
void dfs(long long s) {
  vis[s] = true;
  for (auto &it : adj[s]) {
    if (!vis[it]) {
      dfs(it);
    }
  }
}
void solve(long long tc) {
  cin >> n >> m;
  pw[0] = 1;
  for (long long i = 1; i <= n; i++) {
    pw[i] = 271 * pw[i - 1];
  }
  for (long long i = 1; i <= m; i++) {
    long long x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
    ha[x] += pw[y];
    ha[y] += pw[x];
  }
  long long idx = 1;
  for (long long i = 1; i <= n; i++) {
    if (ha[i] == 0) {
      cout << -1 << '\n';
      return;
    }
    if (ma[ha[i]] == 0) {
      ma[ha[i]] = idx++;
    }
  }
  if ((long long)ma.size() == 3) {
    for (long long i = 1; i <= n; i++) {
      cout << ma[ha[i]] << " ";
    }
  } else {
    cout << -1 << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long cnt = 1;
  long long t = 1;
  while (t--) {
    for (auto blockTime = make_pair(chrono::high_resolution_clock::now(), true);
         blockTime.second;
         fprintf(stderr, "%s: %lld ms\n", "solve",
                 chrono::duration_cast<chrono::milliseconds>(
                     chrono::high_resolution_clock::now() - blockTime.first)
                     .count()),
              fflush(stderr), blockTime.second = false) {
      solve(cnt);
    }
    cnt++;
  }
  return 0;
}
