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;
inline int read() {
  char ch = getchar();
  int nega = 1;
  while (!isdigit(ch)) {
    if (ch == '-') nega = -1;
    ch = getchar();
  }
  int ans = 0;
  while (isdigit(ch)) {
    ans = ans * 10 + ch - 48;
    ch = getchar();
  }
  if (nega == -1) return -ans;
  return ans;
}
vector<int> G[300005];
int n, m;
int ans[300005];
bool vis[300005];
int cnt[5];
signed main() {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int u = read(), v = read();
    G[u].push_back(v), G[v].push_back(u);
  }
  int sum = 1;
  queue<int> q;
  while (!q.empty()) q.pop();
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; ++i) {
    if (!ans[i]) {
      q.push(i);
    }
  }
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    while (!q.empty() && vis[x]) {
      x = q.front();
      q.pop();
    }
    if (vis[x]) break;
    ans[x] = sum;
    int tx = G[x].size();
    for (int i = 0; i < tx; i++) {
      vis[G[x][i]] = 1;
    }
  }
  sum++;
  while (!q.empty()) q.pop();
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; ++i) {
    if (!ans[i]) {
      q.push(i);
    }
  }
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    while (!q.empty() && vis[x]) {
      x = q.front();
      q.pop();
    }
    if (vis[x]) break;
    ans[x] = sum;
    int tx = G[x].size();
    for (int i = 0; i < tx; i++) {
      vis[G[x][i]] = 1;
    }
  }
  sum++;
  while (!q.empty()) q.pop();
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; ++i) {
    if (!ans[i]) {
      q.push(i);
    }
  }
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    while (!q.empty() && vis[x]) {
      x = q.front();
      q.pop();
    }
    if (vis[x]) break;
    ans[x] = sum;
    int tx = G[x].size();
    for (int i = 0; i < tx; i++) {
      vis[G[x][i]] = 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!ans[i]) {
      cout << "-1\n";
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    cnt[ans[i]]++;
  }
  if (!cnt[1] || !cnt[2] || !cnt[3]) {
    printf("-1\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    int tx = G[i].size();
    for (int j = 0; j < tx; j++) {
      int to = G[i][j];
      if (ans[i] == ans[to]) {
        printf("-1\n");
        return 0;
      }
    }
    if (tx != n - cnt[ans[i]]) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  cout << "\n";
  return 0;
}
