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, long long z) {
  long long ret = 1;
  while (y > 0) {
    if (y & 1) ret = (ret * x) % z;
    x = (x * x) % z;
    y >>= 1;
  }
  return ret;
}
const long long N = 1e5 + 5, P1 = 32452867, P2 = 49979687, M1 = 1e9 + 7,
                M2 = 1e9 + 9;
vector<long long> V[N];
long long has[N][2], cc = 0, vis[N];
set<pair<long long, long long> > col1;
void dfs(long long x) {
  vis[x] = 1;
  for (auto u : V[x]) {
    if (!vis[u]) dfs(u);
  }
  return;
}
void getHash(long long x) {
  long long p = P1, mod = M1;
  for (auto u : V[x]) {
    has[x][0] = (has[x][0] + (u * p) % mod) % mod;
    p = (p * p) % mod;
  }
  p = P2, mod = M2;
  for (auto u : V[x]) {
    has[x][1] = (has[x][0] + (u * p) % mod) % mod;
    p = (p * p) % mod;
  }
  col1.insert(make_pair(has[x][0], has[x][1]));
  return;
}
int main() {
  long long n, m;
  scanf("%lld%lld", &n, &m);
  while (m--) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    V[a].push_back(b);
    V[b].push_back(a);
  }
  for (long long i = 1; i <= n; ++i) sort((V[i]).begin(), (V[i]).end());
  for (long long i = 1; i <= n; ++i)
    if (!vis[i]) dfs(i), ++cc;
  if (cc > 1)
    printf("-1\n");
  else {
    for (long long i = 1; i <= n; ++i) getHash(i);
    long long sz = col1.size();
    if (sz != 3)
      printf("-1\n");
    else {
      map<pair<long long, long long>, long long> mk;
      long long cnt = 1;
      for (auto u : col1) mk[u] = cnt++;
      for (long long i = 1; i <= n; ++i)
        printf("%lld ", mk[make_pair(has[i][0], has[i][1])]);
      printf("\n");
    }
  }
  return 0;
}
