#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
//#include<cctype>
#include<climits>
#include<iostream>
#include<string>
#include<vector>
#include<map>
//#include<list>
#include<queue>
#include<deque>
#include<algorithm>
//#include<numeric>
#include<utility>
#include<complex>
//#include<memory>
#include<functional>
#include<cassert>
#include<set>
#include<stack>

const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<int, int> pii;

struct UnionFind {
    vector<int> par;
    int n, cnt;
    UnionFind(const int& x = 0) {init(x);}
    void init(const int& x) {par.assign(cnt=n=x, -1);}
    inline int find(const int& x) {return par[x] < 0 ? x : par[x] = find(par[x]);}
    inline bool same(const int& x, const int& y) {return find(x) == find(y);}
    inline bool unite(int x, int y) {
        if ((x = find(x)) == (y = find(y))) return false;
        --cnt;
        if (par[x] > par[y]) swap(x, y);
        par[x] += par[y];
        par[y] = x;
        return true;
    }
    inline int count() const {return cnt;}
    inline int count(int x) {return -par[find(x)];}
};

const int INF = 1e8;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N, M;
    cin >> N >> M;
    UnionFind uf(N);
    for (int i = 0; i < M; i++) {
        int u, v;
        cin >> u >> v;
        u--; v--;
        uf.unite(u, v);
    }
    vector<int> memo(N);
    for (int i = 0; i < N; i++) memo[uf.find(i)]++;
    vector<int> cnt(N+1);
    for (int i = 0; i < N; i++) if (memo[i]) cnt[memo[i]]++;
    vector<int> ans(N+1, INF);
    ans[0] = 0;
    for (int i = 1; i <= N; i++) {
        for (int k = 1; cnt[i]; k <<= 1) {
            int num = min(cnt[i], k);
            for (int j = N-num*i; j >= 0; j--) {
                int nj = j+num*i;
                ans[nj] = min(ans[nj], ans[j]+num);
            }
            cnt[i] -= num;
        }
    }
    for (int i = 1; i <= N; i++) cout << ((ans[i]==INF) ? -1 : ans[i]-1) << endl;
    return 0;
}