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 maxn = 1e5 + 7;
set<int> G[maxn], G2[maxn];
pair<int, int> edge[maxn * 10];
int n, m;
int vis[maxn];
int vis2[maxn];
int f[maxn];
vector<int> t, t2, t3;
int K[3];
vector<int> s[10];
int getf(int x) {
  if (x == f[x]) return x;
  return f[x] = getf(f[x]);
}
int main() {
  scanf("%d %d", &n, &m);
  int a, b;
  for (int i = 1; i <= n; i++) f[i] = i;
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &a, &b);
    edge[i] = make_pair(a, b);
    f[getf(a)] = getf(b);
    G[a].insert(b);
    G[b].insert(a);
  }
  int k = 0;
  for (int i = 1; i <= n; i++)
    if (i == getf(i)) k++;
  if (k != 1) {
    printf("-1\n");
    return 0;
  }
  for (auto j : G[1]) vis[j] = 1;
  for (int i = 1; i <= n; i++)
    if (vis[i] == 0) t2.push_back(i);
  for (int i = 0; i < m; i++) {
    if (vis[edge[i].first] == 0 && vis[edge[i].second] == 0) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) s[vis[i]].push_back(i);
  for (auto i : s[0]) {
    for (auto j : s[1]) {
      if (G[i].find(j) == G[i].end()) {
        printf("-1\n");
        return 0;
      }
    }
  }
  for (int i = 0; i < m; i++) {
    if (vis[edge[i].first] == 1 && vis[edge[i].second] == 1) {
      G2[edge[i].first].insert(edge[i].second);
      G2[edge[i].second].insert(edge[i].first);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (G2[i].size() != 0) {
      for (auto j : G2[i]) vis2[j] = 1;
      break;
    }
  }
  for (int i = 1; i <= n; i++)
    if (vis[i] == 1 && vis2[i] == 0) t3.push_back(i);
  for (int i = 0; i < m; i++) {
    if (vis[edge[i].first] == 1 && vis2[edge[i].first] == 0 &&
        vis[edge[i].second] == 1 && vis2[edge[i].second] == 0) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++)
    if (vis[i] == 1 && vis2[i] == 0)
      s[2].push_back(i);
    else if (vis[i] == 1 && vis2[i] == 1)
      s[3].push_back(i);
  for (auto i : s[2]) {
    for (auto j : s[3]) {
      if (G2[i].find(j) == G2[i].end()) {
        printf("-1\n");
        return 0;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (vis[i] == 1 && vis2[i] == 1) t.push_back(i);
  }
  for (int i = 0; i < m; i++) {
    if (vis[edge[i].first] == 1 && vis2[edge[i].first] == 1 &&
        vis[edge[i].second] == 1 && vis2[edge[i].second] == 1) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++)
    if (vis[i] == 0)
      K[0]++;
    else {
      if (vis2[i] == 0)
        K[1]++;
      else
        K[2]++;
    }
  for (int i = 0; i < 3; i++)
    if (K[i] == 0) {
      printf("-1\n");
      return 0;
    }
  for (int i = 1; i <= n; i++) {
    if (vis[i] == 0) {
      printf("1 ");
    } else {
      if (vis2[i] == 0)
        printf("2 ");
      else
        printf("3 ");
    }
  }
  return 0;
}
