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 fir[100005], nex[600005], tar[600005], cnt;
int n, m, vis[100005], lab[100005], s[5];
void Read(int &p) {
  p = 0;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar()) p = p * 10 + c - '0';
}
void Add(int a, int b) {
  ++cnt;
  tar[cnt] = b;
  nex[cnt] = fir[a];
  fir[a] = cnt;
}
bool Check(int c) {
  for (int i = 1; i <= n; i++)
    if (lab[i] == c) {
      int sum = 0;
      for (int j = fir[i]; j; j = nex[j]) {
        int v = tar[j];
        if (lab[v] == c) return 0;
        sum++;
      }
      if (sum != n - s[c]) return 0;
    }
  return 1;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int a, b;
    Read(a), Read(b);
    Add(a, b), Add(b, a);
    if (a == 1 || b == 1) vis[a + b - 1] = 1;
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) lab[i] = 1, s[1]++;
  if (!Check(1)) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; i++)
    if (!lab[i]) lab[i] = 2, s[2]++;
  for (int i = 1; i <= n; i++)
    if (lab[i] == 2) {
      for (int j = fir[i]; j; j = nex[j]) {
        int v = tar[j];
        if (lab[v] == 2) lab[v] = 3, s[2]--, s[3]++;
      }
      break;
    }
  if (!Check(2) || !Check(3)) {
    puts("-1");
    return 0;
  }
  if (!s[1] || !s[2] || !s[3]) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; i++) printf("%d%c", lab[i], i == n ? 10 : 32);
}
