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;
struct node {
  int to, next;
} q[600005];
int head[300005], cnt = 1;
int d[300005];
vector<int> v1, v2;
void add(int x, int y) {
  q[cnt].to = y;
  q[cnt].next = head[x];
  head[x] = cnt++;
}
int main() {
  memset(head, -1, sizeof(head));
  int n, m, a, b;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &a, &b);
    v1.push_back(a);
    v2.push_back(b);
    add(a, b);
    add(b, a);
  }
  d[1] = 1;
  int val;
  for (int i = head[1]; i != -1; i = q[i].next) {
    int v = q[i].to;
    d[v] = 2;
    val = v;
  }
  d[val] = 2;
  for (int i = head[val]; i != -1; i = q[i].next) {
    int v = q[i].to;
    if (d[v] == 2) {
      d[v] = 3;
    }
  }
  long long a1 = 0, a2 = 0, a3 = 0;
  for (int i = 1; i <= n; i++) {
    if (d[i] == 0 || d[i] == 1) {
      d[i] = 1;
      a1++;
    }
    if (d[i] == 2) {
      a2++;
    }
    if (d[i] == 3) {
      a3++;
    }
  }
  if (a1 == 0 || a2 == 0 || a3 == 0) {
    printf("-1\n");
    return 0;
  }
  if (a1 * a2 + a1 * a3 + a2 * a3 == m) {
    int flag = 1;
    for (int i = 0; i < m; i++) {
      if (d[v1[i]] == d[v2[i]]) {
        flag = 0;
        break;
      }
    }
    if (!flag) {
      printf("-1\n");
    } else {
      for (int i = 1; i <= n; i++) {
        printf("%d ", d[i]);
      }
      printf("\n");
    }
  } else {
    printf("-1\n");
  }
  return 0;
}
