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 double pi = acos(-1.0);
const double eps = 1e-6;
const unsigned long long hashp = 1e7 + 19;
const long long INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 100;
int col[maxn], head[maxn], cnt;
int du[maxn], p[maxn], sz[maxn], vis[maxn];
struct node {
  int nxt, to;
} e[maxn * 6];
void add(int a, int b) { e[++cnt].nxt = head[a], e[cnt].to = b, head[a] = cnt; }
int find(int x) { return x == p[x] ? x : p[x] = find(p[x]); }
void solve() {
  int n, m;
  scanf("%d%d", &n, &m);
  while (m--) {
    int a, b;
    scanf("%d%d", &a, &b);
    add(a, b);
    add(b, a);
    du[a]++;
    du[b]++;
  }
  for (int i = 1; (i) <= (n); i++) p[i] = i, sz[i] = 1;
  for (int o = 1; (o) <= (3); o++) {
    int now = -1;
    for (int i = 1; (i) <= (n); i++)
      if (!col[i]) {
        now = i;
        break;
      }
    if (now == -1) {
      printf("-1");
      return;
    }
    int u = now;
    col[u] = o;
    for (int i = head[u]; i; i = e[i].nxt) {
      int to = e[i].to;
      vis[to] = 1;
    }
    for (int i = 1; (i) <= (n); i++)
      if (!vis[i] && i != now) {
        if (col[i]) {
          printf("-1");
          return;
        }
        int x = find(i), y = find(u);
        if (x != y) p[x] = y, sz[y] += sz[x];
        col[i] = o;
      }
    memset(vis, 0, sizeof(vis));
  }
  for (int i = 1; (i) <= (n); i++)
    if (!col[i]) {
      printf("-1");
      return;
    }
  for (int i = 1; (i) <= (n); i++) {
    int x = find(i);
    if (sz[x] == n - du[x] && du[x] == du[i])
      ;
    else {
      printf("-1");
      return;
    }
  }
  for (int u = 1; (u) <= (n); u++) {
    for (int i = head[u]; i; i = e[i].nxt) {
      int to = e[i].to;
      if (col[u] == col[to]) {
        printf("-1");
        return;
      }
    }
  }
  for (int i = 1; (i) <= (n); i++) printf("%d ", col[i]);
}
int main() {
  solve();
  return 0;
}
