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 inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline int gc() {
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read() {
  int ret = 0, f = 0;
  char c = gc();
  while (!isdigit(c)) {
    if (c == '-') f = 1;
    c = gc();
  }
  while (isdigit(c)) {
    ret = ret * 10 + c - 48;
    c = gc();
  }
  if (f) return -ret;
  return ret;
}
const int maxn = 3e5 + 10;
struct Edge {
  int v, next;
} edge[maxn << 1];
int head[maxn / 3], tot;
void init() {
  memset(head, -1, sizeof(head));
  tot = 0;
}
void addedge(int u, int v) {
  edge[tot].v = v;
  edge[tot].next = head[u];
  head[u] = tot++;
}
int vis[maxn / 3];
bool book[4];
int main() {
  init();
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = (1); i <= (m); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    addedge(u, v);
    addedge(v, u);
  }
  for (int i = (1); i <= (n); i++) {
    vis[i] = 1;
  }
  for (int i = head[1]; i != -1; i = edge[i].next) {
    int v = edge[i].v;
    vis[v] = 2;
  }
  int t = -1;
  for (int i = (1); i <= (n); i++) {
    if (vis[i] == 2) {
      t = i;
      break;
    }
  }
  if (t == -1) {
    printf("%d\n", t);
    return 0;
  }
  for (int i = head[t]; i != -1; i = edge[i].next) {
    int v = edge[i].v;
    if (vis[v] == 1) continue;
    vis[v] = 3;
  }
  t = -1;
  for (int i = (1); i <= (n); i++) {
    if (vis[i] == 3) {
      t = i;
      break;
    }
  }
  if (t == -1) {
    printf("%d\n", t);
    return 0;
  }
  for (int i = head[t]; i != -1; i = edge[i].next) {
    int v = edge[i].v;
    if (vis[v] == 3) {
      printf("%d\n", -1);
      return 0;
    }
  }
  for (int u = (1); u <= (n); u++) {
    book[1] = book[2] = book[3] = 0;
    book[vis[u]] = 1;
    for (int i = head[u]; i != -1; i = edge[i].next) {
      int v = edge[i].v;
      if (vis[u] == vis[v]) {
        printf("%d\n", -1);
        return 0;
      }
      book[vis[v]] = 1;
    }
    if (!(book[1] && book[2] && book[3])) {
      printf("%d\n", -1);
      return 0;
    }
  }
  int cnt1 = 0, cnt2 = 0, cnt3 = 0;
  for (int i = (1); i <= (n); i++) {
    if (vis[i] == 1)
      cnt1++;
    else if (vis[i] == 2)
      cnt2++;
    else
      cnt3++;
  }
  for (int u = (1); u <= (n); u++) {
    int tmp1 = 0, tmp2 = 0, tmp3 = 0;
    for (int i = head[u]; i != -1; i = edge[i].next) {
      int v = edge[i].v;
      if (vis[v] == 1) tmp1++;
      if (vis[v] == 2) tmp2++;
      if (vis[v] == 3) tmp3++;
    }
    if (vis[u] == 1 && (!(tmp2 == cnt2 && tmp3 == cnt3))) {
      printf("%d\n", -1);
      return 0;
    }
    if (vis[u] == 2 && (!(tmp1 == cnt1 && tmp3 == cnt3))) {
      printf("%d\n", -1);
      return 0;
    }
    if (vis[u] == 3 && (!(tmp2 == cnt2 && tmp1 == cnt1))) {
      printf("%d\n", -1);
      return 0;
    }
  }
  for (int i = (1); i <= (n); i++) {
    printf("%d ", vis[i]);
  }
  return 0;
}
