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;
const int N = 1e5 + 10;
vector<int> G[N];
int n, m, a, b;
int vis[N];
int cnt[3];
bool check() {
  for (int i = 1; i <= n; ++i) {
    for (auto j : G[i]) {
      if (vis[i] == vis[j]) return 0;
    }
  }
  return 1;
}
int main() {
  scanf("%d%d", &n, &m);
  vis[0] = vis[n + 1] = 100;
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  for (int i = 1; i <= n; ++i) {
    G[i].push_back(0);
    G[i].push_back(n + 1);
  }
  vis[1] = 1;
  sort(G[1].begin(), G[1].end());
  for (int i = 0; i + 1 < G[1].size(); ++i) {
    for (int j = G[1][i] + 1; j < G[1][i + 1]; ++j) vis[j] = 1;
  }
  int s = 0;
  for (int i = 1; i <= n; ++i) {
    if (vis[i] != 1) {
      s = i;
      break;
    }
  }
  if (s == 0) {
    puts("-1");
    return 0;
  }
  sort(G[s].begin(), G[s].end());
  vis[s] = 2;
  for (int i = 0; i + 1 < G[s].size(); ++i) {
    for (int j = G[s][i] + 1; j < G[s][i + 1]; ++j) vis[j] = 2;
  }
  for (int i = 1; i <= n; ++i) cnt[vis[i]]++;
  if (1LL * cnt[0] * cnt[1] + 1LL * cnt[1] * cnt[2] + 1LL * cnt[0] * cnt[2] !=
          m ||
      1LL * cnt[0] * cnt[1] * cnt[2] == 0 || !check()) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; ++i) printf("%d%c", vis[i] + 1, " \n"[i == n]);
  return 0;
}
