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;
const int maxn = 1e5 + 1000;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
vector<int> g[maxn];
int col[maxn];
int main(void) {
  int n, m;
  n = read();
  m = read();
  for (int i = 1; i <= m; i++) {
    int u, v;
    u = read();
    v = read();
    g[u].push_back(v);
    g[v].push_back(u);
  }
  if (g[1].size() == 0) {
    printf("-1\n");
    return 0;
  }
  int one = 1;
  col[1] = 1;
  int two = g[1][0];
  col[two] = 2;
  int three = 0;
  for (int i = 1; i <= n; i++) {
    if (i == one || i == two) continue;
    bool f1 = 0;
    bool f2 = 0;
    for (int j = 0; j < g[i].size(); j++) {
      int v = g[i][j];
      if (v == one) f1 = 1;
      if (v == two) f2 = 1;
    }
    if (f1 && f2) {
      three = i;
      col[three] = 3;
      break;
    }
  }
  if (!three) {
    printf("-1\n");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    int f1 = 0;
    int f2 = 0;
    int f3 = 0;
    for (int j = 0; j < g[i].size(); j++) {
      int v = g[i][j];
      if (v == one) f1++;
      if (v == two) f2++;
      if (v == three) f3++;
    }
    if (f1 + f2 + f3 != 2) {
      printf("-1\n");
      return 0;
    }
    if (f1 && f2 && !f3)
      col[i] = 3;
    else if (f1 && f3 && !f2)
      col[i] = 2;
    else if (f2 && f3 && !f1)
      col[i] = 1;
  }
  map<int, int> map1;
  for (int i = 1; i <= n; i++) {
    map1[col[i]]++;
  }
  int num1 = map1[1];
  int num2 = map1[2];
  int num3 = map1[3];
  if (num1 * num2 + num2 * num3 + num3 * num1 != m) {
    printf("-1\n");
    return 0;
  } else {
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < g[i].size(); j++) {
        if (col[g[i][j]] == col[i]) {
          printf("-1");
          return 0;
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      printf("%d ", col[i]);
    }
    printf("\n");
  }
  return 0;
}
