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 N = 1e5;
int n, m, Vdisv, Tt;
int Edu[N + 3];
vector<int> Elnk[N + 3];
int Bbvis[N + 3], Eans[N + 3], Sbvis[N + 3];
void DFS(int u) {
  Bbvis[u] = 1;
  Vdisv++;
  for (auto v : Elnk[u]) {
    if (Bbvis[v]) continue;
    DFS(v);
  }
}
void GFS(int u, int Sb) {
  Sbvis[u] = Sb;
  for (auto v : Elnk[u]) {
    if (Eans[v] != Tt) continue;
    if (Sbvis[v] != -1) {
      if (Sbvis[v] == Sbvis[u])
        printf("-1\n"), exit(0);
      else
        continue;
    }
    GFS(v, !Sb);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    Edu[u]++;
    Edu[v]++;
    Elnk[u].push_back(v);
    Elnk[v].push_back(u);
  }
  DFS(1);
  if (Vdisv != n) {
    printf("-1\n");
    return 0;
  }
  Vdisv = 0;
  memset(Bbvis, 0, sizeof Bbvis);
  for (int i = 1; i <= n; i++)
    if (!Bbvis[Edu[i]]) Vdisv++, Bbvis[Edu[i]] = 1;
  if (Vdisv > 3) {
    printf("-1\n");
    return 0;
  }
  memset(Bbvis, 0, sizeof Bbvis);
  int Vcnt = 0;
  for (int i = 1; i <= n; i++)
    if (!Bbvis[Edu[i]]) {
      Bbvis[Edu[i]] = ++Vcnt;
      Eans[i] = Vcnt;
    } else
      Eans[i] = Bbvis[Edu[i]];
  if (Vdisv == 3) {
    for (int u = 1; u <= n; u++)
      for (auto v : Elnk[u])
        if (Eans[u] == Eans[v]) {
          printf("-1\n");
          return 0;
        }
    for (int i = 1; i <= n; i++) {
      printf("%d", Eans[i]);
      if (i == n)
        printf("\n");
      else
        printf(" ");
    }
    return 0;
  }
  if (Vdisv == 2) {
    memset(Sbvis, -1, sizeof Sbvis);
    Tt = 1;
    for (int i = 1; i <= n && Tt == 1; i++)
      if (Eans[i] != Tt)
        for (auto j : Elnk[i])
          if (Eans[j] == Eans[i]) {
            Tt = 2;
            break;
          }
    for (int i = 1; i <= n; i++)
      if (Eans[i] == Tt && Sbvis[i] == -1) {
        GFS(i, 1);
        break;
      }
    for (int i = 1; i <= n; i++) {
      if (Eans[i] != Tt)
        printf("1");
      else
        printf("%d", Sbvis[i] + 2);
      if (i == n)
        printf("\n");
      else
        printf(" ");
    }
    return 0;
  }
  if (n % 3) printf("-1\n"), exit(0);
  if (Edu[1] != n / 3 * 2) printf("-1\n"), exit(0);
  for (auto i : Elnk[1]) Eans[i] = 2;
  for (int i = 1; i <= n; i++)
    if (Eans[i] == 0)
      Eans[i] = 1;
    else if (Eans[i] == 2)
      for (auto v : Elnk[i])
        if (Eans[v] == 2) Eans[v] = 3;
  for (int i = 1; i <= n; i++) {
    printf("%d", Eans[i]);
    if (i == n)
      printf("\n");
    else
      printf(" ");
  }
  return 0;
}
