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 = 300005;
int n, m;
struct Node {
  int u, v;
  Node *next;
} * h[MAXN], pool[MAXN << 1];
int tot = 0;
inline void addEdge(int u, int v) {
  Node *p = &pool[++tot], *q = &pool[++tot];
  p->u = u;
  p->v = v;
  p->next = h[u];
  h[u] = p;
  q->u = v;
  q->v = u;
  q->next = h[v];
  h[v] = q;
}
int bl[MAXN], size1, size2, size3;
int main() {
  cin >> n >> m;
  int u, v;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &u, &v);
    addEdge(u, v);
  }
  for (int i = 1; i <= n; i++) {
    bool flag = true;
    int s1 = 0, s2 = 0, s3 = 0;
    for (Node *p = h[i]; p; p = p->next) {
      if (bl[p->v] == 1)
        flag = false;
      else if (bl[p->v] == 2)
        s2++;
      else if (bl[p->v] == 3)
        s3++;
    }
    if (flag && size2 == s2 && size3 == s3) {
      bl[i] = 1;
      size1++;
      continue;
    }
    flag = true;
    s1 = s2 = s3 = 0;
    for (Node *p = h[i]; p; p = p->next) {
      if (bl[p->v] == 2)
        flag = false;
      else if (bl[p->v] == 3)
        s3++;
      else if (bl[p->v] == 1)
        s1++;
    }
    if (flag && size3 == s3 && size1 == s1) {
      bl[i] = 2;
      size2++;
      continue;
    }
    flag = true;
    s1 = s2 = s3 = 0;
    for (Node *p = h[i]; p; p = p->next) {
      if (bl[p->v] == 3)
        flag = false;
      else if (bl[p->v] == 1)
        s1++;
      else if (bl[p->v] == 2)
        s2++;
    }
    if (flag && size1 == s1 && size2 == s2) {
      bl[i] = 3;
      size3++;
      continue;
    }
    cout << -1 << endl;
    return 0;
  }
  if (1ll * size1 * size2 * size3 == 0) {
    cout << -1 << endl;
    return 0;
  }
  if (1ll * size1 * size2 + 1ll * size2 * size3 + 1ll * size3 * size1 != m) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) printf("%d ", bl[i]);
  cout << endl;
  return 0;
}
