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;
struct node {
  int uid;
  basic_string<int> str;
} e[100005];
int deg[100005], col[100005], num[3];
int read() {
  int A;
  bool K;
  char C;
  C = A = K = 0;
  while (C < '0' || C > '9') K |= C == '-', C = getchar();
  while (C > '/' && C < ':') A = (A << 3) + (A << 1) + (C ^ 48), C = getchar();
  return (K ? -A : A);
}
inline bool cmp(node _, node __) { return _.str < __.str; }
int main() {
  int n, m, i, a, b, cnt = 0;
  n = read(), m = read();
  for (i = 1; i <= (m); i++) {
    a = read(), b = read();
    e[a].str += b;
    e[b].str += a;
    deg[a]++;
    deg[b]++;
  }
  for (i = 1; i <= (n); i++)
    sort(e[i].str.begin(), e[i].str.end()), e[i].uid = i;
  sort(e + 1, e + n + 1, cmp);
  for (i = 2; i <= (n); i++)
    if (e[i].str == e[i - 1].str)
      col[e[i].uid] = col[e[i - 1].uid];
    else
      col[e[i].uid] = ++cnt;
  if (cnt != 2) printf("-1"), exit(0);
  for (i = 1; i <= (n); i++) num[col[i]]++;
  for (i = 1; i <= (n); i++)
    if (deg[i] != n - num[col[i]]) printf("-1"), exit(0);
  for (i = 1; i <= (n); i++) printf("%d ", col[i] + 1);
  return 0;
}
