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.

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In the second example, it's impossible to make such vertex sets.
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 50;
int n, m, ver[N * 6], nxt[N * 6], head[N], tot, la, col[N], num[3];
void add(int x, int y) {
  ver[++tot] = y;
  nxt[tot] = head[x];
  head[x] = tot;
}
void f() {
  puts("-1");
  exit(0);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1, x, y; i <= m; i++)
    scanf("%d%d", &x, &y), add(x, y), add(y, x);
  for (int i = head[1], y; i; i = nxt[i]) col[la = y = ver[i]] = 1;
  for (int i = head[la], y; i; i = nxt[i])
    if (col[y = ver[i]]) col[y] = 2;
  for (int i = 1; i <= n; i++) num[col[i]]++;
  if (!num[0] || !num[1] || !num[2]) f();
  for (int x = 1; x <= n; x++) {
    int nn = 0;
    for (int i = head[x], y; i; i = nxt[i], nn++)
      if (col[y = ver[i]] == col[x]) f();
    if (nn != n - num[col[x]]) f();
  }
  for (int i = 1; i <= n; i++) printf("%d ", col[i] + 1);
  return 0;
}
