#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------

ll mo,P,R;
int Q;
ll A,B,C;

ll modpow(ll a, ll n = mo-2) {
	ll r=1;
	while(n) r=r*((n%2)?a:1)%mo,a=a*a%mo,n>>=1;
	return r;
}

ll loglog(ll a,ll b,ll mo) { // a^x=b
	map<ll,int> M;
	ll ap=1,ar=modpow(a);
	ll a_sq=1;
	int i,j;
	int sq=0;
	for(i=0;i*i<=mo;i++) {
		M[b*ap%mo]=i, ap=ap*ar%mo;
		sq=i+1;
		a_sq=a_sq*a%mo;
	}
	
	ll c=1;
	for(i=0;;i+=sq) {
		if(M.count(c)) {
			return (mo+M[c]+i)%mo;
		}
		c=c*a_sq%mo;
	}
}

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>P>>R;
	mo=P;
	
	cin>>Q;
	while(Q--) {
		cin>>A>>B>>C;
		ll r=modpow(A);
		A = A*r%mo;
		B = B*r%mo;
		C = C*r%mo;
		
		if(B%2==1) B+=P;
		B/=2;
		C=(P-(C-B*B%P+P)%P)%P;
		
		if(C==0) {
			cout<<(P-B)%mo<<endl;
		}
		else {
			ll a=loglog(R,C,P);
			if(a%2==1) {
				cout<<-1<<endl;
				continue;
			}
			ll D=modpow(R,a/2);
			
			ll X=(-B+D+2*P)%P;
			ll Y=(-B-D+2*P)%P;
			if(X>Y) swap(X,Y);
			cout<<X<<" "<<Y<<endl;
		}
		
	}
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n';
	FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	solve(); return 0;
}