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 N = 100005;
const int M = 300005;
const long long Mo = 1e9 + 7;
int x, y, n, m;
struct E {
  int to, nxt;
} edge[M * 2];
int col[N], idx[N], tot;
long long cnt[4];
bool vis[N];
void addedge(int from, int to) {
  edge[++tot].to = to;
  edge[tot].nxt = idx[from];
  idx[from] = tot;
}
bool linked(int x, int c) {
  memset(vis, 0, sizeof(vis));
  for (int t = idx[x]; t; t = edge[t].nxt) vis[edge[t].to] = 1;
  for (int i = 1; i <= n; i++) {
    if (col[i] && !vis[i]) {
      return 0;
    }
    if (!vis[i]) col[i] = c;
  }
  return 1;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++)
    scanf("%d%d", &x, &y), addedge(x, y), addedge(y, x);
  if (!linked(1, 1)) return printf("-1\n"), 0;
  for (int i = 1; i <= n; i++)
    if (!col[i]) {
      if (!linked(i, 2)) return printf("-1\n"), 0;
      break;
    }
  for (int i = 1; i <= n; i++)
    if (!col[i]) {
      if (!linked(i, 3)) return printf("-1\n"), 0;
      break;
    }
  for (int i = 1; i <= n; i++) {
    if (col[i] == 0) {
      printf("-1\n");
      return 0;
    }
    for (int c = 1; c <= 3; c++) {
      if (col[i] == c) cnt[c]++;
    }
  }
  if (cnt[1] * cnt[2] + cnt[1] * cnt[3] + cnt[3] * cnt[2] == 1ll * m &&
      cnt[1] != 0 && cnt[2] != 0 && cnt[3] != 0) {
    for (int i = 1; i <= n; i++)
      for (int t = idx[i]; t; t = edge[t].nxt) {
        E e = edge[t];
        if (col[i] == col[e.to]) {
          printf("-1");
          return 0;
        }
      }
    for (int i = 1; i < n; i++) printf("%d ", col[i]);
    printf("%d\n", col[n]);
  } else
    printf("-1\n");
  return 0;
}
