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 P = 1e9 + 7;
inline int add(int a, int b) {
  if ((a += b) >= P) a -= P;
  return a;
}
inline int sub(int a, int b) {
  if ((a -= b) < 0) a += P;
  return a;
}
inline int mul(int a, int b) { return 1ll * a * b % P; }
inline int kpow(int a, int b) {
  int r = 1;
  for (; b; b >>= 1, a = mul(a, a)) {
    if (b & 1) r = mul(r, a);
  }
  return r;
}
const int N = 303030;
int uu[N], vv[N];
int belong[N];
vector<int> G[N];
bool vis[N];
int in[N], cnt;
vector<int> a[3];
int b[3];
int n, m;
bool ok = 1;
void dfs(int p) {
  cnt++;
  vis[p] = 1;
  for (int& v : G[p])
    if (!vis[v]) dfs(v);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = (0); i < (m); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--, v--;
    uu[i] = u, vv[i] = v;
    in[v]++;
    in[u]++;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(0);
  if (cnt != n) return 0 * puts("-1");
  set<int> s;
  for (int i = (0); i < (n); i++) s.insert(in[i]);
  if ((int)s.size() > 3) return 0 * puts("-1");
  cnt = 0;
  memset(b, -1, sizeof(b));
  set<int> a[3];
  if ((int)s.size() == 1) {
    for (int i = (0); i < (n); i++) a[0].insert(i);
    for (int i = (0); i < (m); i++) {
      int u = uu[i], v = vv[i];
      if (belong[u] == belong[v]) {
        belong[v] = 1;
        a[1].insert(v);
        a[0].erase(v);
      }
    }
    for (int i = (0); i < (m); i++) {
      int u = uu[i], v = vv[i];
      if (belong[u] == belong[v]) {
        belong[v] = 2;
        a[2].insert(v);
        a[1].erase(v);
      }
    }
    for (int i = (0); i < (m); i++) {
      int u = uu[i], v = vv[i];
      if (belong[u] == belong[v]) {
        ok = 0;
      }
    }
    for (int i = (0); i < (3); i++)
      if (n - (int)a[i].size() != *s.begin()) ok = 0;
  } else if ((int)s.size() == 2) {
    int cnt = 0;
    for (auto sb : s) {
      b[cnt] = sb;
      cnt++;
    }
    for (int i = (0); i < (n); i++)
      for (int j = (0); j < (2); j++)
        if (in[i] == b[j]) belong[i] = j, a[j].insert(i);
    for (int i = (0); i < (m); i++) {
      int u = uu[i], v = vv[i];
      if (belong[u] == belong[v]) {
        a[belong[v]].erase(v);
        a[2].insert(v);
        belong[v] = 2;
        b[2] = in[v];
      }
    }
    for (int i = (0); i < (3); i++)
      if (n - (int)a[i].size() != b[i]) ok = 0;
  } else if ((int)s.size() == 3) {
    int cnt = 0;
    for (auto sb : s) {
      b[cnt] = sb;
      cnt++;
    }
    for (int i = (0); i < (n); i++)
      for (int j = (0); j < (3); j++)
        if (in[i] == b[j]) belong[i] = j, a[j].insert(i);
    for (int i = (0); i < (m); i++) {
      int u = uu[i], v = vv[i];
      if (belong[u] == belong[v]) {
        ok = 0;
      }
    }
    for (int i = (0); i < (3); i++)
      if (n - (int)a[i].size() != b[i]) ok = 0;
  }
  if (!ok)
    puts("-1");
  else
    for (int i = (0); i < (n); i++) printf("%d ", belong[i] + 1);
  return 0;
}
