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;
int n, m;
int ct[100005];
int ans[100005];
vector<int> adjlist[100005];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    adjlist[a].push_back(b);
    adjlist[b].push_back(a);
    ct[a]++, ct[b]++;
  }
  for (int i = 0; i < n; i++) {
    sort(adjlist[i].begin(), adjlist[i].end());
  }
  memset(ans, -1, sizeof(ans));
  vector<int> v[3];
  for (int k = 0; k <= 2; k++) {
    int sel = -1;
    for (int i = 0; i < n; i++) {
      if (ans[i] == -1) {
        sel = i;
        break;
      }
    }
    if (sel == -1) {
      printf("-1");
      return 0;
    }
    ans[sel] = k;
    v[k].push_back(sel);
    int cur = 0;
    for (int i = sel + 1; i < n; i++) {
      while (cur < adjlist[sel].size() && adjlist[sel][cur] < i) cur++;
      if (cur == adjlist[sel].size() || adjlist[sel][cur] > i) {
        ans[i] = k;
        v[k].push_back(i);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (ans[i] == -1) {
      printf("-1");
      return 0;
    }
  }
  for (int i = 0; i < 3; i++) {
    for (auto x : v[i]) {
      for (auto y : adjlist[x]) {
        if (ans[y] == ans[x]) {
          printf("-1");
          return 0;
        }
      }
    }
  }
  long long check = 0;
  check += v[0].size() * v[1].size();
  check += v[1].size() * v[2].size();
  check += v[2].size() * v[0].size();
  if (check != m) {
    printf("-1");
    return 0;
  }
  for (int i = 0; i < n; i++) {
    printf("%d ", ans[i] + 1);
  }
}
