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<set<int>> graph(300000);
vector<pair<int, int>> A;
int ans[300000];
long long f(long long x) { return x * (x - 1) / 2; }
int main() {
  fill(ans, ans + 300000, 2);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int v1, v2;
    cin >> v1 >> v2;
    v1--;
    v2--;
    graph[v1].insert(v2);
    graph[v2].insert(v1);
    A.push_back({v1, v2});
  }
  ans[0] = 0;
  int First = n;
  for (int i = n - 1; i >= 1; i--) {
    if (graph[0].find(i) == graph[0].end()) {
      ans[i] = 0;
    } else {
      First = i;
    }
  }
  int cnt_0 = 0, cnt_1 = 0, cnt_2 = 0;
  if (First != n) {
    for (int i = n - 1; i >= 1; i--) {
      if (graph[First].find(i) == graph[First].end() && ans[i] == 2) {
        ans[i] = 1;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (ans[i] == 0) {
      cnt_0++;
    } else if (ans[i] == 1) {
      cnt_1++;
    } else {
      cnt_2++;
    }
  }
  long long total = f(n) - f(cnt_0) - f(cnt_1) - f(cnt_2);
  if (total != m || cnt_0 == 0 || cnt_1 == 0 || cnt_2 == 0) {
    cout << -1;
    return 0;
  }
  for (auto i : A) {
    if (ans[i.first] == ans[i.second]) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << ans[i] + 1 << ' ';
  }
}
