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>
#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("avx2")
using namespace std;
int col[100005], mark[100005];
vector<int> adj[100005];
bool vis[100005];
void dfs(int root) {
  vis[root] = true;
  for (auto i : adj[root]) {
    if (!vis[i] && mark[i]) {
      col[i] = col[root] ^ 1;
      dfs(i);
    } else if (mark[i] == 1) {
      if (col[i] == col[root]) {
        cout << "-1\n";
        exit(0);
      }
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m, x, y, mm;
  cin >> n >> m;
  mm = m;
  while (m--) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  for (auto i : adj[1]) {
    mark[i] = 1;
  }
  int two = (int)adj[1].size();
  if (two > 1) {
    dfs(adj[1][0]);
  } else {
    cout << "-1\n";
    return 0;
    ;
  }
  int cnt = 0;
  set<int> st;
  for (auto i : adj[1]) {
    st.insert(col[i]);
  }
  if ((int)st.size() != 2) {
    cout << "-1\n";
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (mark[i] == 1) continue;
    int temp = 0;
    for (auto j : adj[i]) {
      if (vis[j] != 1) {
        cout << "-1\n";
        return 0;
      }
      temp++;
    }
    cnt++;
    if (temp != two) {
      cout << "-1\n";
      return 0;
    }
    col[i] = -1;
  }
  int edges[4] = {0};
  for (int i = 1; i <= n; i++) {
    edges[col[i] + 2]++;
  }
  if (edges[1] * edges[2] + edges[2] * edges[3] + edges[1] * edges[3] != mm) {
    cout << "-1\n";
    return 0;
  }
  if (cnt + two != n) {
    cout << "-1 \n";
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    cout << col[i] + 2 << " ";
  }
}
