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 int MAX = 5e5 + 5;
int n, m, mk[MAX], col[MAX];
vector<int> g[MAX];
void dfs(int u) {
  mk[u] = 1;
  for (int v : g[u])
    if (!mk[v]) dfs(v);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  int cc = 0;
  for (int i = 1; i <= n; i++) {
    if (!mk[i]) {
      cc++;
      dfs(i);
    }
  }
  if (cc > 1) {
    printf("-1\n");
    return 0;
  }
  for (int u = 1; u <= n; u++) {
    vector<int> whom(4, 0);
    for (int v : g[u])
      if (col[v] > 0) whom[col[v]] = 1;
    int c = 1;
    while (c <= 3 && whom[c]) c++;
    if (c > 3) {
      printf("-1\n");
      exit(0);
    }
    assert(c >= 1 && c <= 3);
    col[u] = c;
  }
  map<int, int> mp;
  for (int i = 1; i <= n; i++) mp[col[i]]++;
  if ((int)mp.size() != 3) {
    printf("-1\n");
    return 0;
  }
  for (int i = 1; i <= 3; i++) {
    int nxt = (i == 3) ? 1 : i + 1;
    int c = 0;
    for (int u = 1; u <= n; u++) {
      if (col[u] != i) continue;
      for (int v : g[u]) {
        if (col[v] == nxt) c++;
      }
    }
    if (c != 1LL * mp[i] * mp[nxt]) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", col[i]);
  printf("\n");
  return 0;
}
