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>
const int N = 100005;
int n, m, tot[4], a[4][N], col[N];
std::map<std::pair<int, int>, int> ma;
int main() {
  scanf("%d%d", &n, &m);
  if (!m) {
    puts("-1");
    return 0;
  }
  int u, v;
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    ma[std::make_pair(x, y)] = ma[std::make_pair(y, x)] = 1;
    u = x;
    v = y;
  }
  col[u] = 1;
  col[v] = 2;
  a[1][++tot[1]] = u;
  a[2][++tot[2]] = v;
  for (int i = 1; i <= n; i++) {
    if (i == u || i == v) continue;
    if (!ma[std::make_pair(i, u)] && !ma[std::make_pair(i, v)]) {
      puts("-1");
      return 0;
    }
    if (!ma[std::make_pair(i, u)])
      col[i] = 1, a[1][++tot[1]] = i;
    else if (!ma[std::make_pair(i, v)])
      col[i] = 2, a[2][++tot[2]] = i;
    else
      col[i] = 3, a[3][++tot[3]] = i;
  }
  if ((long long)tot[1] * tot[2] + (long long)tot[2] * tot[3] +
              (long long)tot[3] * tot[1] !=
          (long long)m ||
      !tot[3]) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i < 3; i++)
    for (int j = i + 1; j <= 3; j++)
      for (int k = 1; k <= tot[i]; k++)
        for (int l = 1; l <= tot[j]; l++)
          if (!ma[std::make_pair(a[i][k], a[j][l])]) {
            puts("-1");
            return 0;
          }
  for (int i = 1; i <= n; i++) printf("%d ", col[i]);
  return 0;
}
