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;
int d[100000];
vector<int> e[100000];
int find(int x) {
  if (x != d[x]) return d[x] = find(d[x]);
  return x;
}
void join(int x, int y) {
  x = find(x);
  y = find(y);
  d[x] = y;
}
int dp[100000];
int answer[100000], id;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    d[i] = i;
  }
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--;
    y--;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  for (int i = 0; i < n; i++) {
    sort(e[i].begin(), e[i].end());
    int lat = -1;
    for (int y : e[i]) {
      if (lat + 1 < y) {
        dp[lat + 1]++;
        dp[y - 1]--;
        join(i, lat + 1);
      }
      lat = y;
    }
    if (lat + 1 < n) {
      dp[lat + 1]++;
      dp[n - 1]--;
      join(i, lat + 1);
    }
  }
  for (int i = 0; i < n; i++) {
    if (i) dp[i] += dp[i - 1];
    if (dp[i]) {
      join(i, i + 1);
    }
  }
  for (int i = 0; i < n; i++) {
    if (find(i) == i) {
      answer[i] = ++id;
    }
  }
  if (id != 3) {
    puts("-1");
    return 0;
  }
  long long cnt[4] = {};
  for (int i = 0; i < n; i++) {
    answer[i] = answer[find(i)];
    cnt[answer[i]]++;
  }
  if (cnt[1] * cnt[2] + cnt[2] * cnt[3] + cnt[1] * cnt[3] != m) {
    puts("-1");
    return 0;
  }
  for (int i = 0; i < n; i++) {
    printf("%d ", answer[i]);
  }
  return 0;
}
