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;
vector<int> adj[100001];
int col[100001];
int cnt[100001];
int flag;
void dfs(int v) {
  for (int u : adj[v]) {
    if (col[u] != 1) {
      if (col[u] == 0) {
        col[u] = col[v] ^ 1;
        dfs(u);
      } else {
        if (col[u] == col[v]) {
          flag = 0;
        }
      }
    }
  }
  return;
}
int main() {
  int n, m, u, v;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    col[i] = 1;
    cnt[i] = 0;
  }
  for (int i = 0; i < m; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i : adj[1]) {
    col[i] = 0;
    cnt[i]++;
  }
  int cnt1 = 1, cnt2 = 0, cnt3 = 0;
  for (int i = 2; i <= n; i++) {
    if (col[i]) {
      cnt1++;
      for (int j : adj[i]) {
        cnt[j]++;
      }
    }
  }
  flag = 1;
  for (int i = 1; i <= n; i++) {
    if (col[i]) {
      if (cnt[i] != 0) flag = 0;
    } else {
      if (cnt[i] != cnt1) flag = 0;
    }
  }
  if (flag) {
    for (int i = 1; i <= n; i++) {
      if (!col[i]) {
        col[i] = 2;
        dfs(i);
      }
    }
    for (int i = 1; i <= n; i++) {
      if (col[i] == 2) {
        cnt2++;
        for (int j : adj[i]) {
          cnt[j]++;
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      if (col[i] == 0)
        flag = 0;
      else if (col[i] == 1) {
        if (cnt[i] != cnt2) flag = 0;
      } else if (col[i] == 2) {
        if (cnt[i] != cnt1) flag = 0;
      } else {
        cnt3 = 1;
        if (cnt[i] != cnt1 + cnt2) flag = 0;
      }
    }
    if (flag && cnt3) {
      for (int i = 1; i <= n; i++) {
        cout << col[i] << " ";
      }
      cout << "\n";
    } else
      cout << "-1\n";
  } else
    cout << "-1\n";
  return 0;
}
