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 maxn = 1e5 + 10;
vector<int> vc1, vc2, vc3;
vector<int> vc[maxn];
int vis[maxn], v[maxn];
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    for (int i = 0; i <= n; i++) vc[i].clear();
    vc1.clear(), vc2.clear(), vc3.clear();
    memset(vis, 0, sizeof(vis));
    memset(v, 0, sizeof(v));
    for (int i = 0; i < m; i++) {
      int x, y;
      scanf("%d%d", &x, &y);
      vc[x].push_back(y);
      vc[y].push_back(x);
    }
    vc1.push_back(1);
    int ans = 1;
    v[1] = 1;
    if (vc[1].size()) {
      int x = vc[1][0];
      v[x] = 2;
      ans++;
      vc2.push_back(x);
      for (int j = 1; j < vc[1].size(); j++) vis[vc[1][j]]++;
      int ff = 0;
      for (int j = 0; j < vc[x].size(); j++) {
        vis[vc[x][j]]++;
        if (vis[vc[x][j]] == 2) {
          vc3.push_back(vc[x][j]), v[vc[x][j]] = 3, ans++;
          ff = 1;
          break;
        }
      }
      if (!ff) {
        printf("-1\n");
        continue;
      }
      int vx[5];
      int f = 0;
      for (int i = 2; i <= n; i++) {
        if (ans == n) {
          for (int j = 1; j <= n; j++) printf("%d ", v[j]);
          printf("\n");
          f = 1;
          break;
        }
        if (v[i]) continue;
        vx[1] = vx[2] = vx[3] = 0;
        for (int j = 0; j < vc[i].size(); j++) {
          x = vc[i][j];
          vx[v[x]]++;
        }
        if (!vx[1]) {
          if (vx[2] == vc2.size() && vx[3] == vc3.size())
            v[i] = 1, ans++, vc1.push_back(i);
          else {
            f = 1;
            printf("-1\n");
            break;
          }
        } else if (!vx[2]) {
          if (vx[1] == vc1.size() && vx[3] == vc3.size())
            v[i] = 2, ans++, vc2.push_back(i);
          else {
            f = 1;
            printf("-1\n");
            break;
          }
        } else if (!vx[3]) {
          if (vx[1] == vc1.size() && vx[2] == vc2.size())
            v[i] = 3, ans++, vc3.push_back(i);
          else {
            f = 1;
            printf("-1\n");
            break;
          }
        } else {
          f = 1;
          printf("-1\n");
          break;
        }
      }
      if (!f) {
        if (ans == n) {
          for (int j = 1; j <= n; j++) printf("%d ", v[j]);
          printf("\n");
        } else
          printf("-1\n");
      }
    } else
      printf("-1\n");
  }
  return 0;
}
