#pragma GCC optimization ("O3")

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using vec = vector<ll>;
using mat = vector<vec>;
using pll = pair<ll,ll>;

#define INF (1LL<<61)
#define MOD 1000000007LL 
//#define MOD 998244353LL
#define EPS (1e-10)

#define PR(x) cout << (x) << endl
#define PS(x) cout << (x) << " "
#define REP(i,m,n) for(ll (i)=(m),(i_len)=(n);(i)<(i_len);++(i))
#define FORE(i,v) for(auto (i):v)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((ll)(x).size())
#define REV(x) reverse(ALL((x)))
#define ASC(x) sort(ALL((x)))
#define DESC(x) {ASC((x)); REV((x));}
#define BIT(s,i) (((s)>>(i))&1)
#define pb push_back
#define fi first
#define se second

template<class T> inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;}
template<class T> inline int chmax(T& a, T b) {if(a<b) {a=b; return 1;} return 0;}
class mint {
public:
    ll x;
    mint(ll x=0) : x((x%MOD+MOD)%MOD) {}
    mint operator-() const {return mint(-x);}
    mint& operator+=(const mint& a) {if((x+=a.x)>=MOD) x-=MOD; return *this;}
    mint& operator-=(const mint& a) {if((x+=MOD-a.x)>=MOD) x-=MOD; return *this;}
    mint& operator*=(const mint& a) {(x*=a.x)%=MOD; return *this;}
    mint operator+(const mint& a) const {mint b(*this); return b+=a;}
    mint operator-(const mint& a) const {mint b(*this); return b-=a;}
    mint operator*(const mint& a) const {mint b(*this); return b*=a;}
    mint pow(ll t) const {if(!t) return 1; mint a=pow(t>>1); return (t&1?*this*a:a)*a;}
    mint inv() const {return pow(MOD-2);}
    mint& operator/=(const mint& a) {return *this*=a.inv();}
    mint operator/(const mint& a) const {mint b(*this); return b/=a;}
};
istream &operator>>(istream& is, mint& a) {ll t; is>>t; a=t; return is;}
ostream &operator<<(ostream& os, const mint& a) {return os<<a.x;}

ll S;
map<ll,ll> D;

ll modpow(ll a, ll b, ll m)
{
    if(b == 0) return 1;
    ll t = modpow(a, b>>1, m);
    return ((b&1)?a*t%m:t)*t%m;
}

ll gs(ll X, ll Y, ll M)
{
    if(D.count(Y)) return D[Y];
    ll R = modpow(modpow(X, M, M), M-2, M);
    REP(i,1,S+1) {
        Y = Y*R%M;
        if(D.count(Y)) return D[Y]+i*S;
    }
    return -1;
}

void solve(ll P, ll R)
{
    ll A, B, C;
    cin >> A >> B >> C;

    ll D = ((B*B-4*A*C)%P+P)%P;
    if(D == 0) {
        ll x = -B*modpow(2*A, P-2, P); x = (x%P+P)%P;
        PR(x);
    }
    else {
        ll k = gs(R, D, P);
        if(k%2 == 0) {
            ll x1 = modpow(R, k/2, P);
            ll x2 = modpow(R, (k+P-1)/2, P);
            x1 = (-B+x1)*modpow(2*A, P-2, P); x1 = (x1%P+P)%P;
            x2 = (-B+x2)*modpow(2*A, P-2, P); x2 = (x2%P+P)%P;
            if(x1 == x2) PR(x1);
            else if(x1 < x2) PS(x1), PR(x2);
            else PS(x2), PR(x1);
        }
        else PR(-1);
    }
}

int main()
{
    ll P, R, Q;
    cin >> P >> R >> Q;

    ll z = 1;
    S = sqrt(P)+1;
    D[1] = 0;
    REP(i,0,S) {
        z = z*R%P;
        D[z] = i+1;
    }

    REP(i,0,Q) solve(P, R);  

    return 0;
}

/*



*/