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;
vector<set<int> > ad;
int main() {
  int m;
  long long n;
  scanf("%lld%d", &n, &m);
  int se[n], ed[m][2];
  ad.resize(n);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &ed[i][0], &ed[i][1]);
    ed[i][0]--;
    ed[i][1]--;
    ad[ed[i][0]].insert(ed[i][1]);
    ad[ed[i][1]].insert(ed[i][0]);
  }
  memset(se, 0, sizeof se);
  se[0] = 1;
  long long c1, c2;
  c1 = 1;
  for (int i = 1; i < n; i++)
    if (ad[0].count(i) <= 0) {
      se[i] = 1;
      c1++;
    }
  int i = 0;
  while (i < n && se[i] == 1) i++;
  if (i == n) {
    printf("-1\n");
    return 0;
  }
  se[i] = 2;
  c2 = 1;
  for (int j = 1; j < n; j++)
    if (j != i && ad[i].count(j) <= 0)
      if (se[j] == 0) {
        se[j] = 2;
        c2++;
      } else {
        printf("-1\n");
        return 0;
      }
  long long c3 = n - c1 - c2;
  if (c3 <= 0 || c1 * c2 + c2 * c3 + c1 * c3 != m) {
    printf("-1\n");
    return 0;
  }
  for (i = 0; i < m; i++)
    if (se[ed[i][0]] == se[ed[i][1]]) {
      printf("-1\n");
      return 0;
    }
  for (i = 0; i < n; i++) printf("%d\n", se[i] + 1);
  return 0;
}
