#include <iostream>
#include <string>
#include <cmath>
#include <algorithm>
#define task "race"
#define ld long double
#define ll long long
#define ull unsigned long long

using namespace std;
const int sizearr = 100003;

ll k, a[1000006], m;
int n;

inline void fastIOfileinput(){
    ios_base:: sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    if(fopen(task".inp", "r")){
        freopen(task".inp", "r", stdin);
        freopen(task".out", "w", stdout);
    }
    if(fopen(task".in", "r")){
        freopen(task".in", "r", stdin);
        freopen(task".out", "w", stdout);
    }
}

inline void Enter(){
    cin >> k >> n;
    for(int i = 1; i <= n; ++i) cin >> a[i];
}

inline void solve(ll x){
    m = (ll) sqrt( (ld) (2 * k - 2LL + x * x + x) / 2); // Tim ong v lon nhat ma Bessie le dc van toc v roi xuong x + 1 truoc khi di
                                                        //duoc k kilometers
    ll v = (2 * k - 2LL + x * x + x - 2 * m * m) / 2, V = m - x - 1LL; // Nhu vay, thoi gian tam co la v + (v - x - 1)
    m += V + 1LL + (v <= x - 1LL ? (ll)0 : (ll) ceil ( (ld) (v - x + 1LL) * 1.0 / m) ); // Con du mot doan chua di het, ta di not
    // Vi truoc khi den diem cuoi, Bessie da dat van toc x, nen neu nhu quang duong can di con lai ma  <= x - 1 thi se di het ma kcan
    //them chut thoi gian nao, nguoc lai ta di het x, van con 1 quang duong can di, nen ta tinh not
}

int main(){
    fastIOfileinput();
    Enter();
    for(int i = 1; i <= n; ++i){
        solve(a[i]);
        cout << m << (i == n ? "" : "\n");
    }
}