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;
using LL = long long;
const int mo = 1e9 + 7;
LL qpow(LL a, LL b) {
  LL ans = 1;
  a %= mo;
  while (b) {
    if (b & 1) ans = ans * a % mo;
    b >>= 1;
    a = a * a % mo;
  }
  return ans;
}
const int maxn = 3e5 + 5;
int what[maxn];
struct node {
  int a, b;
  bool operator<(const node& _) const {
    if (a != _.a) return a < _.a;
    return b < _.b;
  }
} a[maxn << 1];
bool flag[maxn];
map<pair<int, int>, bool> my;
int A[maxn], B[maxn], C[maxn];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int i, j;
  int cnt = 0;
  for (i = 0; i < m; i++) {
    scanf("%d%d", &a[cnt].a, &a[cnt].b);
    if (a[cnt].a > a[cnt].b) swap(a[cnt].a, a[cnt].b);
    my[{a[cnt].a, a[cnt].b}] = 1;
    cnt++;
    a[cnt] = {a[cnt - 1].b, a[cnt - 1].a};
    cnt++;
  }
  sort(a, a + cnt);
  memset(what, -1, sizeof(what));
  for (i = 0; i < cnt; i++) {
    if (a[i].a == 1) flag[a[i].b] = 1;
  }
  for (i = 1; i <= n; i++)
    if (!flag[i]) what[i] = 1;
  for (i = 1; i <= n; i++) {
    if (what[i] == -1) break;
  }
  if (i == n + 1) {
    printf("-1\n");
    exit(0);
  }
  int now = i;
  memset(flag, 0, sizeof(flag));
  for (i = 0; i < cnt; i++) {
    if (a[i].a == now) {
      flag[a[i].b] = 1;
    }
  }
  for (i = 1; i <= n; i++) {
    if (!flag[i]) {
      if (what[i] != -1) {
        printf("-1\n");
        exit(0);
      }
      what[i] = 2;
    }
  }
  LL num1 = 0, num2 = 0, num3 = 0;
  for (i = 1; i <= n; i++) {
    if (what[i] == 1) {
      A[num1++] = i;
    } else if (what[i] == 2) {
      B[num2++] = i;
    } else
      C[num3++] = i;
  }
  if (!num1 || !num2 || !num3) {
    printf("-1\n");
    exit(0);
  }
  if ((num1 - 1) * (num2 - 1) + (num1 - 1) * (num3 - 1) +
          (num2 - 1) * (num3 - 1) >
      m) {
    printf("-1\n");
    exit(0);
  }
  int ta, tb;
  for (i = 0; i < num1; i++) {
    for (j = 0; j < num2; j++) {
      ta = A[i], tb = B[j];
      if (ta > tb) swap(ta, tb);
      if (!my[{ta, tb}]) {
        printf("-1\n");
        exit(0);
      }
      my[{ta, tb}] = 0;
    }
  }
  for (i = 0; i < num1; i++) {
    for (j = 0; j < num3; j++) {
      ta = A[i], tb = C[j];
      if (ta > tb) swap(ta, tb);
      if (!my[{ta, tb}]) {
        printf("-1\n");
        exit(0);
      }
      my[{ta, tb}] = 0;
    }
  }
  for (i = 0; i < num2; i++) {
    for (j = 0; j < num3; j++) {
      ta = B[i], tb = C[j];
      if (ta > tb) swap(ta, tb);
      if (!my[{ta, tb}]) {
        printf("-1\n");
        exit(0);
      }
      my[{ta, tb}] = 0;
    }
  }
  for (const auto& x : my) {
    if (x.second == 1) {
      printf("-1\n");
      exit(0);
    }
  }
  for (i = 1; i < n; i++) {
    if (what[i] == -1)
      printf("3 ");
    else
      printf("%d ", what[i]);
  }
  if (what[i] == -1)
    printf("3 ");
  else
    printf("%d ", what[i]);
  return 0;
}
