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("unroll-loops", "no-stack-protector", "O3")
using namespace std;
const int N = (int)1e5 + 5;
int fa[N], vis[N], cc[N], n, m, d = 0, cnt = 0;
vector<int> G[N], group[N];
vector<pair<int, int> > E;
inline int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
inline void bfs(int s) {
  static queue<int> q;
  q.push(s);
  fa[s] = find(s + 1);
  group[cnt].push_back(s);
  cc[s] = cnt;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (auto v : G[x]) vis[v] = x;
    for (int i = find(1); i <= n; i = find(i + 1)) {
      if (vis[i] != x)
        fa[i] = find(i + 1), q.push(i), group[cnt].push_back(i), cc[i] = cnt;
    }
  }
  ++cnt;
}
void init() {
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
    E.emplace_back(a, b);
  }
  for (int i = 1; i <= n + 1; ++i) fa[i] = i;
}
void solve() {
  for (int i = 1; i <= n; i = find(i + 1)) bfs(i);
  if (cnt == 3) {
    int count[3] = {0};
    for (int i = 1; i <= n; ++i) count[cc[i]]++;
    if ((long long)count[0] * count[1] + (long long)count[1] * count[2] +
            (long long)count[2] * count[0] !=
        m) {
      cout << "-1\n";
      return;
    }
    for (int i = 1; i <= n; ++i) cout << cc[i] + 1 << ' ';
  } else
    cout << "-1\n";
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  init();
  solve();
}
