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 comp {
  bool operator()(int a, int b) { return a > b; }
};
vector<int> node[100005];
int main() {
  int n, m, a, b, ans[100005];
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &a, &b);
    node[a].push_back(b);
    node[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) ans[i] = 1;
  int size = (int)node[1].size();
  for (int i = 0; i <= size - 1; i++) ans[node[1][i]] = 2;
  if (size > 0) {
    int curNode = node[1][0];
    size = (int)node[curNode].size();
    for (int i = 0; i <= size - 1; i++)
      if (ans[node[curNode][i]] == 2) ans[node[curNode][i]] = 3;
  }
  int nodeCnt[4] = {};
  for (int i = 1; i <= n; i++) nodeCnt[ans[i]]++;
  bool valid = (nodeCnt[1] > 0 && nodeCnt[2] > 0 && nodeCnt[3] > 0);
  if (valid) {
    for (int i = 1; i <= n; i++) {
      int cnt[4] = {};
      size = (int)node[i].size();
      for (int j = 0; j <= size - 1; j++) cnt[ans[node[i][j]]]++;
      for (int j = 1; j <= 3; j++)
        if ((j == ans[i] && cnt[j] != 0) ||
            (j != ans[i] && cnt[j] != nodeCnt[j]))
          valid = false;
      if (!valid) break;
    }
  }
  if (!valid)
    printf("-1\n");
  else {
    for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
    printf("\n");
  }
  return 0;
}
