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> g[100001];
int a[300001], b[300001], ans[100001];
bool vis[100001], vis2[100001];
int p[100001];
int root(int a) { return p[a] == a ? a : (p[a] = root(p[a])); }
void merge(int a, int b) {
  a = root(a), b = root(b);
  if (a != b) p[a] = b;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) p[i] = i;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a[i], &b[i]);
    g[a[i]].push_back(b[i]);
    g[b[i]].push_back(a[i]);
  }
  int l = 0;
  for (int i = 0; i < g[1].size(); i++) {
    int v = g[1][i];
    vis[v] = 1;
    l = v;
  }
  for (int i = 0; i < g[l].size(); i++) {
    int v = g[l][i];
    if (vis[v]) vis2[v] = 1;
  }
  vector<int> x, y, z;
  for (int i = 1; i <= n; i++) {
    if (!vis[i])
      x.push_back(i);
    else if (vis[i] && !vis2[i])
      y.push_back(i);
    else
      z.push_back(i);
  }
  for (int i = 1; i < x.size(); i++) {
    merge(x[0], x[i]);
  }
  for (int i = 1; i < y.size(); i++) {
    merge(y[0], y[i]);
  }
  for (int i = 1; i < z.size(); i++) {
    merge(z[0], z[i]);
  }
  bool ok = 1;
  for (int i = 1; i <= m; i++) {
    if (root(a[i]) == root(b[i])) ok = 0;
  }
  if (x.size() + y.size() + z.size() != n || x.size() == 0 || y.size() == 0 ||
      z.size() == 0)
    ok = 0;
  for (int i = 0; i < x.size(); i++) {
    int u = x[i];
    if (g[u].size() != y.size() + z.size()) ok = 0;
  }
  for (int i = 0; i < y.size(); i++) {
    int u = y[i];
    if (g[u].size() != x.size() + z.size()) ok = 0;
  }
  for (int i = 0; i < z.size(); i++) {
    int u = z[i];
    if (g[u].size() != x.size() + y.size()) ok = 0;
  }
  if (!ok)
    printf("-1\n");
  else {
    for (int i = 0; i < x.size(); i++) {
      ans[x[i]] = 1;
    }
    for (int i = 0; i < y.size(); i++) {
      ans[y[i]] = 2;
    }
    for (int i = 0; i < z.size(); i++) {
      ans[z[i]] = 3;
    }
    for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  }
}
