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;
void boost() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const int N = 1e6;
const long long inf = 1e9 + 7;
const long double PI = acos(-1);
long long bin_pow(long long x, long long y) {
  long long res = 1, base = x;
  while (y) {
    if (y & 1) {
      res *= base;
      res %= inf;
    }
    base *= base;
    base %= inf;
    y /= 2;
  }
  return res;
}
int n, m;
int color[N];
vector<int> g[N];
vector<int> nums[5];
bool used[N];
void dfs(int v, int curcolor) {
  used[v] = 1;
  color[v] = curcolor;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (used[to]) {
      continue;
    }
    dfs(to, 5 - curcolor);
  }
}
int main() {
  boost();
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int i = 0; i < g[1].size(); i++) {
    color[g[1][i]] = 2;
  }
  for (int i = 1; i <= n; i++) {
    if (!color[i]) {
      color[i] = 1;
      used[i] = 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (color[i] == 1) {
      continue;
    }
    if (used[i]) {
      continue;
    }
    dfs(i, 2);
  }
  for (int i = 1; i <= n; i++) {
    nums[color[i]].push_back(i);
  }
  if (nums[1].size() * nums[2].size() + nums[1].size() * nums[3].size() +
          nums[2].size() * nums[3].size() !=
      m) {
    cout << -1;
    return 0;
  }
  if (!nums[1].size() || !nums[2].size() || !nums[3].size()) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < g[i].size(); j++) {
      if (color[i] == color[g[i][j]]) {
        cout << -1;
        return 0;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << color[i] << " ";
  }
}
