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.

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In the second example, it's impossible to make such vertex sets.
#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> adj[(int)1e5 + 1];
vector<int> ans;
map<int, int> toone;
map<int, int> totwo;
vector<int> vis;
void dfs(int x) {
  if (vis[x]) return;
  vis[x]++;
  for (auto i : adj[x]) dfs(i);
}
int main() {
  cin >> n;
  ans.assign(n + 1, 0);
  vis.assign(n + 1, 0);
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
    if (x == 1 || y == 1) {
      toone[x]++;
      toone[y]++;
    }
  }
  ans[1] = 1;
  for (int i = 1; i <= n; i++)
    if (toone[i] == 0) ans[i] = 1;
  if (adj[1].empty()) {
    cout << -1;
    return 0;
  }
  int two = adj[1][0];
  ans[adj[1][0]] = 2;
  for (int i = 1; i <= n; i++) {
    int x = count(adj[i].begin(), adj[i].end(), two);
    totwo[i] = x;
  }
  for (int i = 1; i <= n; i++) {
    if (totwo[i] == 0 && ans[i] != 1)
      ans[i] = 2;
    else if (totwo[i] == 0 && ans[i] == 1) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 1; i <= n; i++)
    if (ans[i] == 0) ans[i] = 3;
  for (int i = 1; i <= n; i++) {
    for (auto j : adj[i])
      if (ans[i] == ans[j]) {
        cout << -1;
        return 0;
      }
  }
  dfs(1);
  for (int i = 1; i <= n; i++)
    if (vis[i] == 0) {
      cout << -i;
      return 0;
    }
  int a[3] = {0};
  for (int i = 1; i < n + 1; i++)
    if (ans[i] == 1)
      a[0]++;
    else if (ans[i] == 2)
      a[1]++;
    else
      a[2]++;
  for (int i = 1; i < n + 1; i++) {
    int total = 0;
    for (int j = 0; j < 3; j++) {
      if (j != ans[i] - 1) total += a[j];
    }
    if (adj[i].size() != total) {
      cout << -1;
      return 0;
    }
  }
  if (a[0] == 0 || a[1] == 0 || a[2] == 0) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
}
