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;
struct Node {
  int cnt, id, ans;
};
Node e[100005];
int tl[900005], nx[900005], a[900005], ln;
int st[100005];
int book[100005];
int ans[5], len;
bool cmp(Node x, Node y) { return x.cnt < y.cnt; }
bool cmp2(Node x, Node y) { return x.id < y.id; }
int q[100005], head = 1, tail, in[100005];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) tl[i] = ++ln, e[i].id = i;
  for (int j = 1; j <= m; j++) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[tl[x] = nx[tl[x]] = ++ln] = y;
    a[tl[y] = nx[tl[y]] = ++ln] = x;
    e[x].cnt++, e[y].cnt++;
  }
  sort(e + 1, e + n + 1, cmp);
  ln = 0;
  for (int i = 1; i <= n; i++) {
    if (!book[e[i].cnt]) {
      st[++ln] = e[i].cnt;
      book[e[i].cnt] = true;
    }
  }
  if (ln > 3) return !printf("-1");
  if (ln == 3) {
    int li = 1;
    for (int i = 1; i <= n; i++) {
      if (e[i].cnt != e[i - 1].cnt && i != 1) {
        ans[++len] = i - li;
        li = i;
      }
    }
    ans[++len] = n + 1 - li;
    if (ans[1] * ans[2] + ans[2] * ans[3] + ans[3] * ans[1] != m)
      return !printf("-1");
    for (int i = 1; i <= ans[1]; i++) e[i].ans = 1;
    for (int i = ans[1] + 1; i <= ans[1] + ans[2]; i++) e[i].ans = 2;
    for (int i = ans[1] + ans[2] + 1; i <= ans[1] + ans[2] + ans[3]; i++)
      e[i].ans = 3;
    sort(e + 1, e + n + 1, cmp2);
    for (int i = 1; i <= n; i++)
      for (int j = nx[i]; j; j = nx[j])
        if (e[a[j]].ans == e[i].ans) return !printf("-1");
    for (int i = 1; i <= n; i++) printf("%d ", e[i].ans);
    return 0;
  }
  if (ln == 2) {
    int li = 1;
    for (int i = 1; i <= n; i++) {
      if (e[i].cnt != e[i - 1].cnt && i != 1) {
        ans[++len] = i - li;
        li = i;
      }
    }
    ans[++len] = n + 1 - li;
    if (ans[1] == st[2]) {
      for (int i = ans[1] + 1; i <= ans[1] + ans[2]; i++) e[i].ans = 3;
      if (ans[1] & 1) return !printf("-1");
      if (st[1] - ans[2] == ans[1] / 2) {
        e[1].ans = 1;
        q[++tail] = e[1].id;
        in[e[1].id] = true;
        sort(e + 1, e + n + 1, cmp2);
        while (head <= tail) {
          int i = q[head++];
          for (int j = nx[i]; j; j = nx[j]) {
            if (e[a[j]].ans == 3) continue;
            if (e[a[j]].ans == e[i].ans) return !printf("-1");
            e[a[j]].ans = 3 - e[i].ans;
            if (!in[a[j]]) q[++tail] = a[j], in[a[j]] = true;
          }
        }
        for (int i = 1; i <= n; i++)
          for (int j = nx[i]; j; j = nx[j])
            if (e[a[j]].ans == e[i].ans) return !printf("-1");
        for (int i = 1; i <= n; i++) printf("%d ", e[i].ans);
        return 0;
      }
    }
    if (ans[2] == st[1]) {
      for (int i = 1; i <= ans[1]; i++) e[i].ans = 3;
      if (ans[2] & 1) return !printf("-1");
      if (st[2] - ans[1] == ans[2] / 2) {
        e[n].ans = 1;
        q[++tail] = e[n].id;
        in[e[n].id] = true;
        sort(e + 1, e + n + 1, cmp2);
        while (head <= tail) {
          int i = q[head++];
          for (int j = nx[i]; j; j = nx[j]) {
            if (e[a[j]].ans == 3) continue;
            if (e[a[j]].ans == e[i].ans) return !printf("-1");
            e[a[j]].ans = 3 - e[i].ans;
            if (!in[a[j]]) q[++tail] = a[j], in[a[j]] = true;
          }
        }
        for (int i = 1; i <= n; i++)
          for (int j = nx[i]; j; j = nx[j])
            if (e[a[j]].ans == e[i].ans) return !printf("-1");
        for (int i = 1; i <= n; i++) printf("%d ", e[i].ans);
        return 0;
      }
    }
  }
  if (ln == 1) {
    int li = 1;
    for (int i = 1; i <= n; i++) {
      if (e[i].cnt != e[i - 1].cnt && i != 1) {
        ans[++len] = i - li;
        li = i;
      }
    }
    ans[++len] = n + 1 - li;
    if (2 * ans[1] != 3 * st[1]) return !printf("-1");
    e[1].ans = 1;
    q[++tail] = e[1].id;
    in[e[1].id] = true;
    sort(e + 1, e + n + 1, cmp2);
    while (head <= tail) {
      int i = q[head++];
      for (int j = nx[i]; j; j = nx[j]) {
        if (e[a[j]].ans == e[i].ans) {
          if (e[i].ans == 1)
            return !printf("-1");
          else
            continue;
        }
        e[a[j]].ans = 3 - e[i].ans;
        if (!in[a[j]]) q[++tail] = a[j], in[a[j]] = true;
      }
    }
    head = 1;
    tail = 0;
    memset(in, 0, sizeof(in));
    for (int i = 1; i <= n; i++)
      if (e[i].ans == 2) {
        q[++tail] = i;
        in[i] = true;
        break;
      }
    while (head <= tail) {
      int i = q[head++];
      for (int j = nx[i]; j; j = nx[j]) {
        if (e[a[j]].ans == 1) continue;
        e[a[j]].ans = 5 - e[i].ans;
        if (!in[a[j]]) q[++tail] = a[j], in[a[j]] = true;
      }
    }
    for (int i = 1; i <= n; i++)
      for (int j = nx[i]; j; j = nx[j])
        if (e[a[j]].ans == e[i].ans) return !printf("-1");
    for (int i = 1; i <= n; i++) printf("%d ", e[i].ans);
    return 0;
  }
  return !printf("-1");
}
