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#include <bits/stdc++.h>
#include <bits/extc++.h>

using namespace std;
using namespace __gnu_pbds;

#define DEBUG
#ifdef DEBUG
template<typename T1,typename T2>auto& operator<<(ostream&o,pair<T1,T2>a){return o<<"("<<a.first<<", "<<a.second<<")";}
template<typename T,size_t...I>void pt(ostream&o,T t,index_sequence<I...>){o<<"(";(...,(o<<(I?", ":"")<< get<I>(t)));o<<")";}
template<typename...A>auto& operator<<(ostream&o,tuple<A...>t){pt(o,t,index_sequence_for<A...>{});return o;}
template<typename T,typename O>auto& operator<<(O&o,T a){o<<"{";for(auto b:a)o<<b<<", ";return o<<"}";}
#define db(x...) cerr << "\033[92m" << "[" #x "]: ", [](auto... args) { ((cerr << args << ", "),...) << "\033[0m" << "\n"; }(x)
#else
#define db(...)
#endif

#define sz(x) ((int)(x).size()) 
#define all(x) (x).begin(), (x).end()
#define F first
#define S second

template<class T>
using iset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
using ll = long long;
using ld = long double;
using pll = pair<ll,ll>;
using vi = vector<int>;

mt19937 mrand(random_device{}());
ll rnd(ll l, ll r) { return l + mrand() % (r - l + 1);}

ll n, d;

struct PairHash {
    size_t operator()(const pair<ll,ll>& p) const {
        return hash<ll>()(p.first) ^ (hash<ll>()(p.second) << 32);
    }
};

int main() {
    cin.tie(0)->sync_with_stdio(0); cin.exceptions(cin.failbit);
    
    cin >> d >> n;
    vector<ll> v(n), ans(n);
    for(ll i = 0 ; i < n ; ++i) {
        cin >> v[i]; v[i]--;
    }



    ll k = 11000;
    for(ll i = 2 ; i <= 20'000 ; ++i) {
        if(max((ll)((ld)i/log(i)), 2*d/i) < max((ll)((ld)k/log(k)), 2*d/k)) k = i;
    }
    ll c = 2*d/k;
    k += 7; c += 7;

    vector<int> highest(d+7);
    for(ll i = 2 ; i < d+7 ; ++i) highest[i] = i;
    for(ll i = 2 ; i < d+7 ; ++i) {
        if(highest[i] != i) continue;
        for(ll j = 2*i ; j < d+7 ; j += i) {
            highest[j] = i;
        }
    }

    vector<ll> cnt(n+7), cnt2(n+7);
    vector<vector<int>> mp(k+7);
    for(ll i = 2 ; i < k+7 ; ++i) mp[i].resize(i);
    gp_hash_table<ll, ll> mp2;
    set<ll> active;
    ll mx = 0, mx2 = 0, acc = 0;

    vector<ll> inv(c*(c+1)/2 + 7, 0);
    for(ll i = 2; (i-1)*i/2 < c*(c+1)/2 + 7; i++) inv[(i-1)*i/2] = i;

    vector<ll> distances;
    for(ll i = 2 ; i <= k ; ++i) {
        if(highest[i] == i) distances.emplace_back(i);
    }

    vector<int> t(d+2, 0);
    cc_hash_table<ll, vector<tuple<ll,ll,ll>>> dist;

    for(ll i = 0 ; i < n ; ++i) {
        if(active.count(v[i])) {
            acc--;
            for(auto ds : distances) {
                int& cn = mp[ds][v[i]%ds];
                cnt[cn]--;
                if(cn-1 > 0) cnt[cn-1]++;
                if(mx > 0 && cnt[mx] == 0) mx--;
                cn--;
            }
            for(auto [hi, q, tq] : dist[v[i]]) {
                if(t[q] != tq) continue;
                ll& cn = mp2[hi*(d+7)+v[i]%hi];
                cnt2[cn]--;
                if(cn-1 > 0) cnt2[cn-1]++;
                if(mx2 > 0 && cnt2[mx2] == 0) mx2--;
                cn--;
            }
            t[v[i]]++;
            dist.erase(v[i]);
            active.erase(v[i]);
        } else {
            acc++;
            for(auto ds : distances) {
                int& cn = mp[ds][v[i]%ds];
                cnt[cn+1]++;
                if(cn > 0) cnt[cn]--;
                if(cn+1 > mx) mx = cn+1;
                cn++;
            }
            auto it = active.lower_bound(v[i]);
            for(ll j = 0 ; j < c/2 && it != active.begin() ; ++j) --it;
            for(ll j = 0 ; j < c && it != active.end() ; ++j, ++it) {
                ll q = *it, hi = highest[abs(v[i]-q)];
                if(hi <= k) continue;
                ll cn = mp2[hi*(d+7)+v[i]%hi];
                cnt2[cn+1]++;
                if(cn > 0) cnt2[cn]--;
                if(cn+1 > mx2) mx2 = cn+1;
                mp2[hi*(d+7)+v[i]%hi]++;
                dist[v[i]].emplace_back(hi, q, t[q]);
                dist[q].emplace_back(hi, v[i], t[v[i]]);
            }
            active.emplace(v[i]);
        }
        ans[i] = max({min(2ll, acc), mx, inv[mx2]});
    }


    for(ll i = 0 ; i < n ; ++i) {
        cout << ans[i] << "\n";
    }

    return 0;
}