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 long long MOD = 1000000007;
const int INF = -1;
const int MAXN = 1e5 + 59;
template <typename var = long long>
var fpow(var a, var b, var m) {
  var ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return ret;
}
int deg[MAXN];
int n, m;
bool ans;
map<int, int> mp;
pair<int, int> ed[MAXN * 3];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0, u, v; i < m; ++i) {
    cin >> u >> v;
    deg[u]++;
    deg[v]++;
    ed[i] = make_pair(u, v);
  }
  for (int i = 1; i <= n; ++i) {
    mp[deg[i]]++;
  }
  if (mp.size() == 3) {
    int sum = 0;
    int id = 0;
    for (auto &mi : mp) {
      sum += mi.first;
      mi.second = ++id;
    }
    if (sum == n + n) {
      ans = true;
      for (int i = 1; i <= n; i++) {
        deg[i] = mp[deg[i]];
      }
      for (int i = 0; i < m; i++) {
        if (deg[ed[i].first] == deg[ed[i].second]) {
          ans = false;
          break;
        }
      }
    }
  } else if (mp.size() == 2) {
    int sum = 0;
    for (auto &mi : mp) {
      sum += mi.first;
    }
    int a = mp.begin()->first;
    int b = sum - a;
    int numa = 0;
    int numb = 0;
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
      if (deg[i] == b) {
        numb++;
      } else {
        numa++;
      }
    }
    if (a + a + b == n + n && numb % 2 == 0) {
      ans = true;
      for (int i = 1; i <= n; i++) {
        if (deg[i] == b) {
          deg[i] = 1;
        } else if (cnt < numa / 2) {
          deg[i] = 2;
          cnt++;
        } else {
          deg[i] = 3;
        }
      }
    } else if (a + b + b == n + n && numa % 2 == 0) {
      ans = true;
      for (int i = 1; i <= n; i++) {
        if (deg[i] == a) {
          deg[i] = 1;
        } else if (cnt < numb / 2) {
          deg[i] = 2;
          cnt++;
        } else {
          deg[i] = 3;
        }
      }
    }
  } else if (mp.size() == 1) {
    if (mp.begin()->first * 3 == n + n) {
      ans = true;
      m = n / 3;
      for (int i = 1; i * 3 <= n; i++) {
        deg[i] = 1;
        deg[i + n / 3] = 2;
        deg[i + n / 3 + n / 3] = 3;
      }
    }
  }
  if (ans) {
    for (int i = 1; i <= n; i++) cout << deg[i] << " \n"[i == n];
  } else {
    cout << "-1\n";
  }
  return 0;
}
